Representation Theory of Groups

Home , Class function, Cuspidal representation, Matrix coefficient, Regular representation, Trivial representation, Unitary representation

MAT 445/1196 - INTRODUCTION TO REPRESENTATION THEORY

CHAPTER 1 – Representation Theory of Groups - Algebraic Foundations 1.1 Basic definitions, Schur’s Lemma 1.2 Tensor products 1.3 Unitary representations 1.4 Characters of finite-dimensional representations CHAPTER 2 – Representations of Finite Groups 2.1 Unitarity, complete reducibility, orthogonality relations 2.2 Character values as algebraic integers, degree of an irreducible representation divides the order of the group 2.3 Decomposition of finite-dimensional representations 2.4 Induced Representations, Frobenius reciprocity, and Frobenius character formula

CHAPTER 3 – Representations of SL2(Fq) CHAPTER 4 – Representations of Finite Groups of Lie Type CHAPTER 5 – Topological Groups, Representations, and Haar Measure 5.1 Topological spaces 5.2 Topological groups 5.3 General linear groups and matrix groups 5.4 Matrix Lie groups 5.5 Finite-dimensional representations of topological groups and matrix Lie groups 5.6 Groups of t.d. type 5.7 Haar measure on locally compact groups 5.8 Discrete series representations 5.9 Parabolic subgroups and representations of reductive groups CHAPTER 6 – Representations of Compact Groups 6.1 Examples of compact groups 6.2 Finite-dimensional representations of compact groups 6.3 The Peter-Weyl Theorem 6.4 Weyl’s character formula

1 CHAPTER 1

Representation Theory of Groups - Algebraic Foundations

1.1. Basic definitions, Schur’s Lemma We assume that the reader is familiar with the fundamental concepts of abstract group theory and linear algebra. A representation of a group G is a homomorphism from G to the group GL(V ) of invertible linear operators on V , where V is a nonzero complex vector space. We refer to V as the representation space of π. If V is finite-dimensional, we say that π is finite-dimensional, and the degree of π is the dimension of V . Otherwise, we say that π is infinite-dimensional. If π is one-dimensional, then V ' C and we view π as a homomorphism from G to the multiplicative group of nonzero complex numbers. In the above definition, G is not necessarily finite. The notation (π, V ) will often be used when referring to a representation. Examples: (1) If G is a group, we can define a one-dimensional representation of G by π(g) = 1, g ∈ G. This representation is called the trivial representation of G. (2) Let G = R and z ∈ C. The function t 7→ ezt defines a one-dimensional representation of G.

If n is a positive integer and C is the field of complex numbers, let GLn(C) denote the group of invertible n × n matrices with entries in C. If (π, V ) is a finite-dimensional representation of G, then, via a choice of ordered basis β for V , the operator π(g) ∈ GL(V ) is identified with the element [π(g)]β of GLn(C), where n is the degree of π. Hence we may view a finite-dimensional representation of G as a homomorphism from G to the group GLn(C). Examples:

(1) The self-representation of GLn(C) is the n-dimensional representation deﬁned by π(g) = g.

(2) The function g 7→ det g is a one-dimensional representation of GLn(C). (3) Let V be a space of functions from G to some complex vector space. Suppose that V

has the property that whenever f ∈ V , the function g0 7→ f(g0g) also belongs to V

for all g ∈ G. Then we may deﬁne a representation (π, V ) by (π(g)f)(g0) = f(gg0), f ∈ V , g, g0 ∈ G. For example, if G is a ﬁnite group, we may take V to be the space of all complex-valued functions on G. In this case, the resulting representation is called the right regular representation of G. Let (π, V ) be a representation of G. A subspace W of V is stable under the action of G, or G-invariant, if π(g)w ∈ W for all g ∈ G and w ∈ W . In this case, denoting the

2 restriction of π(g) to W by π |W (g), (π |W ,W ) is a representation of G, and we call it a subrepresentation of π (or a subrepresentation of V ). 0 If W ⊂ W are subrepresentations of π, then each π |W (g), g ∈ G, induces an in- 0 0 vertible linear operator πW/W 0 (g) on the quotient space W/W , and (πW/W 0 , W/W ) is a representation of G, called a subquotient of π. In the special case W = V , it is called a quotient of π. A representation (π, V ) of G is ﬁnitely-generated if there exist ﬁnitely many vectors v1, . . . , vm ∈ V such that V = Span{ π(g)vj | 1 ≤ j ≤ m, g ∈ G }. A representation (π, V ) of G is irreducible if {0} and V are the only G-invariant subspaces of V . If π is not irreducible, we say that π is reducible.

Suppose that (πj,Vj), 1 ≤ j ≤ `, are representations of a group G. Recall that an element of the direct sum V = V1 ⊕ · · · ⊕ V` can be represented uniquely in the form v1 + v2 + ··· + v`, where vj ∈ Vj. Set

π(g)(v1 + ··· + v`) = π1(g)v1 + ··· + π`(g)v`, g ∈ G, vj ∈ Vj, 1 ≤ j ≤ `.

This defines a representation of G, called the direct sum of the representations π1, . . . , π`, sometimes denoted by π1 ⊕ · · · ⊕ π`. We may define infinite direct sums similarly. We say that a representation π is completely reducible (or semisimple) if π is (equivalent to) a direct sum of irreducible representations. Lemma. Suppose that (π, V ) is a representation of G. (1) If π is finitely-generated, then π has an irreducible quotient. (2) π has an irreducible subquotient. Proof. For (1), consider all proper G-invariant subspaces W of V . This set is nonempty and closed under unions of chains (uses finitely-generated). By Zorn’s Lemma, there is a maximal such W . By maximality of W , πV/W is irreducible. Part (2) follows from part (1) since of v is a nonzero vector in V , part (1) says that if W = Span{ π(g)v | g ∈ G }, then π |W has an irreducible quotient. qed Lemma. Let (π, V ) be a finite-dimensional representation of G. Then there exists an irreducible subrepresentation of π.

Proof. If V is reducible, there exists a nonzero G-invariant proper subspace W1 of V . If

π |W1 is irreducible, the proof is complete. Otherwise, there exists a nonzero G-invariant subspace W2 of W1. Note that dim(W2) < dim(W1) < dim(V ). Since dim(V ) < ∞, this process must eventually stop, that is there exist nonzero subspaces Wk ( Wk−1 ( ··· (

W1 ( V , where π |Wk is irreducible. qed Lemma. Let (π, V ) be a representation of G. Assume that there exists an irreducible subrepresentation of π. The following are equivalent:

3 (1) (π, V ) is completely reducible. (2) For every G-invariant subspace W ⊂ V , there exists a G-invariant subspace W 0 such that W ⊕ W 0 = V . Proof. Assume that π is completely reducible. Without loss of generality, π is reducible. Let W be a proper nonzero G-invariant subspace of V . Consider the set of G-invariant subspaces U of V such that U ∩W = {0}. This set is nonempty and closed under unions of chains, so Zorn’s Lemma implies existence of a maximal such U. Suppose that W ⊕U 6= V . Since π is completely reducible, there exists some irreducible subrepresentation U 0 such that U 0 6⊂ W ⊕U. By irreduciblity of U 0, U 0 ∩(W ⊕U) = {0}. This contradicts maximality of U. Suppose that (2) holds. Consider the partially ordered set of direct sums of families of P irreducible subrepresentations: α Wα = ⊕αWα. Zorn’s Lemma applies. Let W = ⊕αWα be the direct sum for a maximal family. By (2), there exists a subrepresentation U such that V = W ⊕ U. If U 6= {0}, according to a lemma above, there exists an irreducible subquotient: U ⊃ U1 ⊃ U2 such that πU1/U2 is is irreducible. By (2), W ⊕ U2 has a G-invariant complement U3: V = W ⊕ U2 ⊕ U3. Now

U3 ' V/(W ⊕ U2) = (W ⊕ U)/(W ⊕ U2) ' U/U2 ⊃ U1/U2.

Identifying πU1/U2 with an irreducible subrepresentation π |U4 of π |U3 , we have W ⊕ U4 contradicting maximality of the family Wα. qed Lemma. Subrepresentations and quotient representations of completely reducible representations are completely reducible. Proof. Let (π, V ) be a completely reducible representation of G. Suppose that W is a proper nonzero G-invariant subspace of W . Then, according to the above lemma, there exists a G-invariant subspace U of V such that V = W ⊕ U. It follows that the subrepresentation π |W is equivalent to the quotient representation πV/U . Therefore it suﬃces to prove that any quotient representation of π is completely reducible.

Let πV/U be an arbitrary quotient representation of π. We know that π = ⊕α∈I πα, where I is some indexing set, and each πα is irreducible. Let pr : V → V/U be the canonical map. Then V/U = pr(V ) = ⊕α∈I pr(Vα). Because pr(Vα) is isomorphic to a quotient of Vα (pr(Vα) ' Vα/ker (pr | Vα)) and πα is irreducible, we have that pr(Vα) is either 0 or irreducible. Hence πV/U is completely reducible. qed Exercises: (1) Show that the self-representation of GLn(C) is irreducible. 1 t (2) Verify that π : t 7→ deﬁnes a representation of , with space 2, that is a 0 1 R C two-dimensional representation of R. Show that there is exactly one one-dimensional

4 subrepresentation, hence π is not completely reducible. Prove that the restriction of π to the unique one-dimensional invariant subspace W is the trivial representation,

and the quotient representation πV/W is the trivial representation.

If (π1,V1) and (π2,V2) are representations of a group G, a linear transformation A : V1 → V2 intertwines π1 and π2 if Aπ1(g)v = π2(g)Av for all v ∈ V1 and g ∈ G. The notation

Hom G(π1, π2) or Hom G(V1,V2) will be used to denote the set of linear transformations from V1 to V2 that intertwine π1 and π2. Two representations (π1,V1) and (π2,V2) of a group G are said to be equivalent (or isomorphic) whenever Hom G(π1, π2) contains an isomorphism, that is, whenever there exists an invertible linear tranformation A : V1 → V2 that intertwines π1 and π2. In this case, we write π1 ' π2. It is easy to check that the notion of equivalence of representations deﬁnes an equivalence relation on the set of representations of G. It follows from the deﬁnitions that if π1 and π2 are equivalent representations, then π1 is irreducible if and only if π2 is irreducible. More generally, π1 is completely reducible if and only if π2 is completely reducible.

Lemma. Suppose that (π1,V1) and (π2,V2) are ﬁnite-dimensional representations of G. Then the following are equivalent:

(1) π1 and π2 are equivalent.

(2) dim V1 = dim V2 and there exist ordered bases β1 and β2 of V1 and V2, respectively,

such that [π1(g)]β1 = [π2(g)]β2 for all g ∈ G.

Proof. Assume (1). Fix ordered bases γ1 for V1 and γ2 for V2. Via these bases, identifying any invertible operator in Hom G(π1, π2) as a matrix A in GLn(C), we have

−1 [π1(g)]γ1 = A [π2(g)]γ2 A, ∀ g ∈ G.

Let β1 = γ1. Because A ∈ GLn(C), there exists an ordered basis β2 of V2 such that A is the change of basis matrix from β2 to γ2. With these choices of β1 and β2, (2) holds.

Now assume that (2) holds. Let A be the unique linear transformation from V1 to V2 which maps the jth vector in β1 to the jth vector in β2. qed A representation (π, V ) of G has a (ﬁnite) composition series if there exist G-invariant subspaces Vj of V such that

{0} ( V1 ( ··· ( Vr = V

each subquotient πVj+1/Vj , 1 ≤ j ≤ r − 1, is irreducible. The subquotients πVj+1/Vj are called the composition factors of π. Lemma. Let (π, V ) be a ﬁnite-dimensional represntation of G. Then π has a composition series. Up to reordering and equivalence, the composition factors of π are unique. Proof left as an exercise.

5 Schur’s Lemma. Let (π1,V1) and (π2,V2) be irreducible representations of G. Then any

nonzero operator in Hom G(π1, π2) is an isomorphism.

Proof. If Hom G(π1, π2) = {0} there is nothing to prove, so assume that it is nonzero. Suppose that A ∈ Hom G(π1, π2) is nonzero. Let g ∈ G and v2 ∈ A(V1). Writing v2 =

A(v1), for some v1 ∈ V1, we have π2(g)v2 = Aπ(g)v1 ∈ A(V1). Hence A(V1) is a nonzero

G-invariant subspace of V2. By irreducibility of π2, we have A(V1) = V2.

Next, let W be the kernel of A. Let v1 ∈ W . Then A(π1(g)v1) = π2(g)(A(v1)) = π1(g)0 = 0 for all g ∈ G. Hence W is a G-invariant proper subspace of V1. By irreducibility

of π1, W = {0}. qed Corollary. Let (π, V ) be a ﬁnite-dimensional irreducible representation of G. Then Hom G(π, π) consists of scalar multiples of the identity operator, that is, Hom G(π, π) ' C.

Proof. Let A ∈ Hom G(π, π). Let λ ∈ C be an eigenvalue of A (such an eigenvalue exists, since V is ﬁnite-dimensional and C is algebraically closed). It is easy to see that A − λI ∈ Hom G(π, π). But A − λI is not invertible. By the previous lemma, A = λI. qed Corollary. If G is an abelian group, then every irreducible ﬁnite-dimensional representation of G is one-dimensional.

Proof left as an exercise.

Exercise: Prove that an irreducible representation of the cyclic group of order n > 1, k 2πimk/n with generator g0, has the form g0 7→ e for some m ∈ { 0, 1, . . . , n − 1 }. (Here i is a complex number such that i2 = −1 and π denotes the area of a circle of radius one).

Let (π, V ) be a representation of G.A matrix coefficient of π is a function from G to C of the form g 7→ λ(π(g)v), for some fixed v ∈ V and λ in the dual space V ∨ of linear functionals on V . Suppose that π is finite-dimensional. Choose an ordered basis ∨ ∨ β = { v1, . . . , vn } of V . Let β = { λ1, . . . , λn } be the basis of V which is dual to β: λj(vi) = δij, 1 ≤ i, j ≤ n. Define a function aij : G → C by [π(g)]β = (aij(g))1≤i,j≤n. Pn Then it follows from π(g)vi = `=1 ai`(g)v` that aij(g) = λj(π(g)vi), so aij is a matrix coefficient of π. If g ∈ G and λ ∈ V ∨, define π∨(g)λ ∈ V ∨ by (π∨(g)λ)(v) = λ(π(g−1)v), v ∈ V . Then (π∨,V ∨) is a representation of G, called the dual (or contragredient) of π.

Exercises: (1) Let (π, V ) be a ﬁnite-dimensional representation of G. Choose β and β∨ as above. ∨ −1 t Show that [π (g)]β∨ = [π(g )]β, for all g ∈ G. Here the superscript t denotes transpose. (2) Prove that if (π, V ) is ﬁnite-dimensional then π is irreducible if and only if π∨ is irreducible.

6 (3) Determine whether the self-representation of GLn(R) (restrict the self-representation of GLn(C) to the subgroup GLn(R)) is equivalent to its dual. (4) Prove that a ﬁnite-dimensional representation of a ﬁnite abelian group is the direct sum of one-dimensional representations.

1.2. Tensor products

Let (πj,Vj) be a representation of a group Gj, j = 1, 2. Recall that V1 ⊗V2 is spanned by elementary tensors, elements of the form v1 ⊗ v2, v1 ∈ V1, v2 ∈ V2. We can deﬁne a representation π1 ⊗ π2 of the direct product G1 × G2 by setting

(π1 ⊗ π2)(g1, g2)(v1 ⊗ v2) = π1(g1)v1 ⊗ π2(g2)v2, gj ∈ Gj, vj ∈ Vj, j = 1, 2, and extending by linearity to all of V1 ⊗ V2. The representation π1 ⊗ π2 of G1 × G2 is called the (external or outer) tensor product of π1 and π2. Of course, when π1 and π2 are

ﬁnite-dimensional, the degree of π1 ⊗ π2 is equal to the product of the degrees of π1 and

π2.

Lemma. Let (πj,Vj) and Gj, j = 1, 2 be as above. Assume that each πj is ﬁnite- dimensional. Then π1 ⊗ π2 is an irreducible representation of G1 × G2 if and only if π1 and π2 are both irreducible.

Proof. If π1 or π2 is reducible, it is easy to see that π1 ⊗ π2 is also reducible.

Assume that π1 is irreducible. Let n = dim V2. Let

n Hom G1 (π1, π1) = Hom G1 (π1, π1) ⊕ · · · ⊕ Hom G1 (π1, π1),

n n and π1 = π1 ⊕ · · · ⊕ π1, where each direct sum has n summands. Then Hom G1 (π1, π1) ' n Hom G1 (π1, π1 ), where the isomorphism is given by A1⊕· · ·⊕An 7→ B, with B(v) = A1(v)⊕

· · · ⊕ An(v). By (the corollary to) Schur’s Lemma, Hom G1 (π1, π1) ' C. Irreducibility of π1 guarantees that given any nonzero v ∈ V1, V1 = Span{ π1(g1)v | g1 ∈ G1 }, and this implies surjectivity. n Because V2 ' C and C ' Hom G1 (π1, π1), we have

n (i) V2 ' Hom G1 (π1, π1 ⊗ 1 ),

n n where π1 ⊗ 1 is the representation of G1 on V1 ⊗ V2 deﬁned by (π1 ⊗ 1 )(g1)(v1 ⊗ v2) =

π1(g1)v1 ⊗ v2, v1 ∈ V1, v2 ∈ V2. (Note that this representation can be identiﬁed with the restriction of π1 ⊗ π2 to the subgroup G1 × {1} of G1 × G2). If m is a positive integer, then

m m (ii) V1 ⊗ Hom G1 (π1, π1 ) → V1 v ⊗ A 7→ A(v)

7 is an isomorphism. Next, we can use (i) and (ii) to show that

{ G1 − invariant subspaces of V1 ⊗ V2 } ↔ { C − subspaces of V2 }

V1 ⊗ W ← W n X →Hom G1 (π1,X) ⊂ Hom G1 (π1, π1 ⊗ 1 ) = V2

As any (G1 × G2)-invariant subspace X of V1 ⊗ V2 is also a G1-invariant subspace, we have

X = V1 ⊗ W for some complex subspace W of V2. If X 6= {0} and π2 is irreducible, then

Span{ (π1 ⊗ π2)(1, g2)X | g2 ∈ G2 } = V1 ⊗ Span{ π2(g2)W | g2 ∈ G2 } = V1 ⊗ V2.

But G1 × G2-invariance of X then forces X = V1 ⊗ V2. It follows that if π1 and π2 are irreducible, then π1 ⊗ π2 is irreducible (as a representation of G1 × G2). qed

Proposition. Let (π, V ) be an irreducible ﬁnite-dimensional representation of G1 × G2.

Then there exist irreducible representations π1 and π2 of G1 and G2, respectively, such that π ' π1 ⊗ π2.

0 0 Proof. Note that π1(g1)v = π((g1, 1))v, g1 ∈ G1, v ∈ V , and π2(g2)v = π((1, g2))v, g2 ∈ G2, v ∈ V , deﬁne representations of G1 and G2, respectively. Choose a nonzero 0 G1-invariant subspace V1 such that π1 |V1 is an irreducible representation of G1. Let v0 be a nonzero vector in V1. Let

0 V2 = Span{ π2(g2)v0 | g2 ∈ G2 }.

0 Then V2 is G2-invariant and π2 := π2 |V2 is a representation of G2, which might be reducible.

Deﬁne A : V1 ⊗V2 → V as follows. Let v1 ∈ V1 and v2 ∈ V2. Then there exist complex (j) Pm (j) numbers cj and elements g1 ∈ G1 such that v1 = j=1 cjπ1(g1 )v0, as well as complex (`) Pn (`) numbers b` and elements g2 ∈ G2 such that v2 = `=1 b`π2(g2 ). Set

m n X X (j) (`) A(v1 ⊗ v2) = cjb`π(g1 , g2 )v0. j=1 `=1

(j) (`) (j) (`) (`) (j) Now π(g1 , g2 )v0 = π1(g1 )π2(g2 )v0 = π2(g2 )π1(g1 )v0. Check that the map A is well-deﬁned, extending to a linear transformation from V1 ⊗ V2 to V . Also check that

A ∈ Hom G1×G2 (V1 ⊗ V2,V ). Because A(v0 ⊗ v0) = v0, we know that A is nonzero. Combining G1 × G2-invariance of A(V1 ⊗ V2) with irreducibility of π, we have A(V1 ⊗ V2) = V . If A also happens to be one-to-one, then we have π1 ⊗ π2 ' π.

8 Suppose that A is not one-to-one. Then Ker A is a G1 × G2-invariant subspace of

V1 ⊗ V2. In particular, Ker A is a G1-invariant subspace of V1 ⊗ V2. Using irreducibility of π1 and arguing as in the previous proof, we can conclude that Ker A = V1 ⊗ W for some complex subspace W of V2. We have an equivalence of the representations (π1 ⊗

π2)(V1⊗V2)/Ker A and π of G1 × G2. To ﬁnish the proof, we must show that the quotient representation (π1 ⊗ π2)V/(V1⊗W ) is a tensor product. If v1 ∈ V1 and v2 ∈ V2, deﬁne

B(v1 ⊗ (v1 + W )) = v1 ⊗ v2 + V1 ⊗ W.

This extends by linearity to a map from V1 ⊗(V2/W ) to the quotient space (V1 ⊗V2)/(V1 ⊗

W ) and it is a simple matter to check that B is an isomorphism and B ∈ Hom G1×G1 (π1 ⊗

(π2)V2/W , (π1 ⊗ π2)(V1⊗V2)/(V1⊗W ). The details are left as an exercise. qed

If (π1,V1) and (π2,V2) are representations of a group G, then we may form the tensor

product representation π1 ⊗π2 of G×G and restrict to the subgroup δG = { (g, g) | g ∈ G } of G×G. This restriction is then a representation of G, also written π1 ⊗π2. It is called the

(inner) tensor product of π1 and π2. Using inner tensor products gives ways to generate

new representations of a group G. However, it is important to note that even if π1 and π2 are both irreducible, the inner tensor product representation π1 ⊗π2 of G can be reducible.

Exercise: Let π1 and π2 be ﬁnite-dimensional irreducible representations of a group G. Prove that the trivial representation of G occurs as a subrepresentation of the (inner) ∨ tensor product representation π1 ⊗ π2 of G if and only if π2 is equivalent to the dual π1 of π.

1.3. Unitary representations

Suppose that (π, V ) is a representation of G. If V is a ﬁnite-dimensional inner product space and there exists an inner product h·, ·i on V such that

hπ(g)v1, π(g)v2i = hv1, v2i, ∀ v1, v2 ∈ V, g ∈ G.

then we say that π is a unitary representation. If V is infinite-dimensional, we say that π is pre-unitary if such an inner product exists, and if V is complete with respect to the norm induced by the inner product (that is, V is a Hilbert space), then we say that π is unitary. Now assume that π is finite-dimensional. Recall that if T is a linear operator on V , the adjoint T ∗ of T is defined by < T (v), w >=< v, T ∗(w) > for all v, w ∈ V . Note that π is unitary if and only if each operator π(g) satisfies π(g)∗ = π(g)−1, g ∈ G. Let n be a positive integer. Recall that if A is an n × n matrix with entries in C, the adjoint A∗ of A is just A∗ = tA¯.

9 Lemma. If (π, V ) is a finite-dimensional unitary representation of G and β is an orthonor- ∗ −1 mal basis of V , then [π(g)]β = [π(g)]β . Proof. Results from linear algebra show that if T is a linear operator on V and β is an ∗ ∗ ∗ −1 orthonormal basis of V , then [T ]β = [T ]β. Combining this with π(g) = π(g) , g ∈ G, proves the lemma. qed Exercises: (1) If (π, V ) is a representation, form a new vector space V¯ as follows. As a set, V = V¯ , and V¯ has the same vector addition as V . If c ∈ C and v ∈ V¯ , set c · v =cv ¯ , wherec ¯ is the complex conjugate of c andcv ¯ is the scalar multiplication in V . If g ∈ G, and v ∈ V¯ ,π ¯(g)v = π(g)v. Show that (¯π, V¯ ) is a representation of V . (2) Assume that (π, V ) is a finite-dimensional unitary representation. Prove that π∨ ' π¯. Lemma. Let W be a subspace of V , where (π, V ) is a unitary representation of G. Then W is G-invariant if and only if W ⊥ is G-invariant. Proof. W is G-invariant if and only if π(g)w ∈ W for all g ∈ G and w ∈ W if and only if hπ(g)w, w⊥i = 0 for all w ∈ W , w⊥ ∈ W ⊥ and g ∈ G if and only if hw, π(g−1)w⊥i = 0 for all w ∈ W , w⊥ ∈ W ⊥ and g ∈ G, if and only if W ⊥ is G-invariant. qed Corollary. A finite-dimensional unitary representation is completely reducible. Lemma. Suppose that (π, V ) is a finite-dimensional unitary represetantion of G. Let W be a proper nonzero G-invariant subspace of V , and let PW be the orthogonal projection of V onto W . Then PW commutes with π(g) for all g ∈ G. Proof. Let w ∈ W and w⊥ ∈ W ⊥. Then

⊥ ⊥ ⊥ PW π(g)(w + w ) = PW π(g)w + PW π(g)w = π(g)w + 0 = π(g)PW (w + w ). qed Lemma. Let (π, V ) be a ﬁnite-dimensional unitary representation of G. Then π is irreducible if and only if Hom G(π, π) ' C (every operator which commutes with all π(g)’s is a scalar multiple of the identity operator). Proof. One direction is simply the corollary to Schur’s Lemma (using irreducibility of π). For the other, if π is reducible, and W is a proper nonzero G-invariant subspace of V ,

Then PW ∈ Hom G(π, π) and PW is not a scalar multiple of the identity operator. qed

Suppose that (π1,V1) and (π2,V2) are representations of G and V1 and V2 are complex

inner product spaces, with inner products h·, ·i1 and h·, ·i2, respectively. Then π1 and π2 are unitarily equivalent if there exists an invertible linear operator A : V1 → V2 such that

hAv, Awi2 = hv, wi1 for all v and w ∈ V1 and A ∈ Hom G(π1, π2).

10 Lemma. Let (π1,V1) and (π2,V2) be ﬁnite-dimensional unitary representations of G.

Then π1 ' π2 if and only if π1 and π2 are unitarily equivalent.

Proof. Assume that π1 ' π2. Let A : V1 → V2 be an isomorphism such that A ∈ ∗ ∗ Hom G(π1, π2). Recall that the adjoint A : V2 → V1 is deﬁned by the condition hA v2, v1i1 =

hv2, Av1i2 for all v1 ∈ V1 and v2 ∈ V2. By assumption, we have

−1 (i) π1(g) = A π2(g)A, ∀ g ∈ G.

∗ ∗ ∗ ∗ −1 Taking adjoints, we have π1(g) = A π2(g) (A ) for all g ∈ G. Since πj is unitary, we ∗ −1 −1 have πj(g) = πj(g ). Replacing g by g, we have

∗ ∗ −1 (ii) π1(g) = A π2(g)(A ) , ∀ g ∈ G.

Expressing π2(g) in terms of π1(g) using (i), we can rewrite (ii) as

∗ −1 ∗ −1 π1(g) = A Aπ1(g)A (A ) , ∀ g ∈ G, or −1 ∗ ∗ π1(g) A Aπ1(g) = A A, ∀ g ∈ G.

Now A∗A is positive definite (that is, self-adjoint and having positive (real) eigenvalues), −1 and so has a unique positive definite square root, say B. Note that π1(g) Bπ1(g) is ∗ ∗ −1 also a square root of A A and it is positive definite, using π1(g) = π1(g) . Hence −1 π1(g) Bπ1(g) = B for all g ∈ G. Writing A in terms of the polar decomposition, we have A = UB, with B as above, and with U an isomorphism from V1 → V2 such that hUv, Uwi2 = hv, wi1 for all v and w ∈ V1. Next, note that

−1 −1 −1 π2(g) = UBπ1(g)B U = Uπ1(g)U , ∀ g ∈ G.

Hence U ∈ Hom G(π1, π2), and π1 and π2 are unitarily equivalent. qed

1.4. Characters of ﬁnite-dimensional representations

Let (π, V ) be a ﬁnite-dimensional representation of a group G. The function g 7→ tr π(g) from G to C is called the character of π. We use the notation χπ(g) = tr π(g). Note that we can use any ordered basis of V to compute χπ(g), since the trace of an operator depends only on the operator itself. Note that if π were inﬁnite-dimensional, the operator π(g) would not have a trace.

11 Lemma. Let (π, V ) be a ﬁnite-dimensional representation of G. 0 (1) If π ' π, then χπ = χπ0 .

(2) The function χπ is constant on conjugacy classes in G. ∨ −1 (3) Let π be the representation dual to π. Then χπ∨ (g) = χπ(g ), g ∈ G. −1 (4) If π is unitary, then χπ(g ) = χπ(g), g ∈ G. (5) Suppose that (π, V ) has a composition series {0} ( V1 ( ··· ( Vr = V , with composition factors π ,π , . . . , π (see page 5). Then χ = χ + χ + ··· + V1 V2/V1 Vr /Vr−1 π πV1 πV2/V1 χπ . Vr /Vr−1

(6) The character χπ1⊗···⊗πr of a tensor product of ﬁnite-dimensional representations π1, . . . , πr of G1,...,Gr, respectively, is given by

χπ1⊗···⊗πr (g1, . . . , gr) = χπ1 (g1)χπ2 (g2) ··· χπr (gr), g1 ∈ G1, . . . , gr ∈ Gr.

Proof. By an earlier result, if π0 ' π, then π0 and π have the same matrix realization (for some choice of bases). Part (1) follows immediately. Note that

−1 −1 χπ(g1gg1 ) = tr (π(g1)π(g)π(g1) ) = tr π(g) = χπ(g), g, g1 ∈ G.

Recall that if β is an ordered basis of V and β∨ is the basis of V ∨ dual to β, then t ∨ −1 [π(g)]β = [π (g )]β∨ . This implies (3). −1 Suppose that π is unitary. Let β be an orthonormal basis of V . Then [π(g )]β = ∗ t [π(g)]β = [π(g)] implies part (4). For (5), it is enough to do the case r = 2. Let β be an ordered basis for V1. Extend β to an ordered basis γ for V2 = V . Letγ ˙ be the ordered basis for V2/V1 which is the image of γ under the canonical map V → V2/V1. Then it is easy to check that [π(g)]γ is equal to

[π | (g)] ∗ V1 β . 0 [πV2/V1 (g)]γ˙

For (6), it is enough to do the case r = 2. Let β = { v1, . . . , vn} and γ = { w1, . . . , wm } be ordered bases of V1 and V2, respectively. Then

{ vj ⊗ w` | 1 ≤ j ≤ n, 1 ≤ ` ≤ m } is an ordered basis of V1 ⊗ V2. Let aij(g1) be the ijth entry of [π1(g1)]β, g1 ∈ G1, and let bij(g2) be the ijth entry of [π2(g2)]γ , g2 ∈ G2. We have

π1(g1)vj = a1j(g1)v1 + a2j(g1)v2 + ··· + anj(g1)vn, g1 ∈ Gr

π2(g2)w` = b1`(g2)w1 + b2`(g2)w2 + ··· + bm`(g2)wm, g2 ∈ G2.

12 Hence n m X X π1(g1)vj ⊗ π2(g2)w` = atj(g1)bs`(g2)(vt ⊗ ws), t=1 s=1 and, as the coeﬃcient of vj ⊗ w` on the right side equals ajj(g1)b``(g2), we have

n m X X χπ1⊗π2 (g1, g2) = ajj(g1)b``(g2) = χπ1 (g1)χπ2 (g2), g1 ∈ G1, g2 ∈ G2. j=1 `=1

Example: The converse to part (1) is false. Consider the Example (2) on page 4. We have χπ(t) = 2 for all t ∈ R. Now take π0 ⊕ π0, where π0 is the trivial representation of R. This clearly has the same character as π, though π0 ⊕ π0 is not equivalent to π. In many cases, for example, if G is ﬁnite, or compact, two irreducible ﬁnite-dimensional representations having the same character must be equivalent.

13 CHAPTER 2

Representations of Finite Groups

In this chapter we consider only ﬁnite-dimensional representations.

2.1. Unitarity, complete reducibility, orthogonality relations

Theorem 1. A representation of a finite group is unitary. c Proof. Let (π, V ) be a (finite- dimensional) representation of a finite group G = { g1, g2, . . . gn }. Let h·, ·i1 be any inner product on V . Set

n X hv, wi = hπ(gj)v, π(gj)wi1, v, w ∈ V. j=1

Then it is clear from the deﬁnition that

hπ(g)v, π(g)wi = hv, wi, v, w ∈ V, g ∈ G.

Pn Note that if v ∈ V , then hv, vi = j=1hπ(gj)v, π(gj)vi1 and if v 6= 0, then π(gj)v 6= 0 for all j implies hπ(gj)v, π(gj)vi1 > 0 for all j. Hence v 6= 0 implies hv, vi > 0. The other properties of inner product are easy to verify for h·, ·i, using the fact that h·, ·i1 is an inner product, and each π(gj) is linear. The details are left as an exercise. qed

The following is an immediate consequence of Theorem 1 and a result from Chapter I stating that a ﬁnite-dimensional unitary representation is completely reducible.

Theorem 2. A representation of a ﬁnite group is completely reducible.

Example. Let G be a finite group acting on a finite set X. Let V be a complex vector space having a basis { vx1 , . . . , vxm } indexed by the elements x1, . . . , xm of X. If g ∈ G, let π(g) be the operator sending vxj to vg·xj , 1 ≤ j ≤ m. Then (π, V ) is a representation of G, called the permutation representation associated with X. Let A(G) be the set of complex-valued functions on G. Often A(G) is called the group algebra of G - see below. Let RA be the (right) regular representation of G on the space A(G): Given f ∈ A(G) and g ∈ G, RA(g)f is the function defined by

(RA(g)f)(g0) = f(g0g), g0 ∈ G. Note that RA is equivalent to the the permutation representation associated to the set X = G. Let LA be the left regular representation of G on the −1 space A(G): Given f ∈ A(G) and g ∈ G, LA(g)f is deﬁned by (LA(g)f)(g0) = f(g g0), ˙ ˙ −1 g0 ∈ G. It is easy to check that the operator f 7→ f, where f(g) = f(g ), is a unitary

equivalence in HomG(RA,LA).

14 If f1, f2 ∈ A(G), the convolution f1 ∗ f2 of f1 with f2 is deﬁned by X −1 (f1 ∗ f2)(g) = f1(gg0 )f2(g0), g ∈ G. g0∈G With convolution as multiplication, A(G) is an algebra. It is possible to study the representations of G in terms of A(G)-modules. We deﬁne an inner product on A(G) as follows:

−1 X hf1, f2i = |G| f1(g)f2(g), f1, f2 ∈ A(G). g∈G

Theorem 3 (Orthogonality relations for matrix coeﬃcients). Let (π1,V1) and i (π2,V2) be irreducible (unitary) representations of G. Let ajk(g) be the matrix entries of the matrix of πi(g) relative to a ﬁxed orthonormal basis of Vi, i = 1, 2 (relative to an inner

product which makes πi unitary). Then 1 2 (1) If π1 6' π2, then hajk, a`mi = 0 for all j, k, ` and m. 1 1 (2) hajk, a`mi = δj`δkm/n1, where n1 = dim V1. −1 P −1 Proof. Let B be a linear transformation from V2 to V1. Then A := |G| g∈G π1(g)Bπ2(g) 0 is also a linear transformation from V2 to V1. Let g ∈ G. Then 0 −1 X 0 −1 −1 X −1 0 0 π1(g )A = |G| π1(g g)Bπ2(g ) = |G| π1(g)Bπ2(g g ) = Aπ2(g ). g∈G g∈G

Hence A ∈ HomG(π2, π1), Let ni = dim Vi, i = 1, 2. Letting bj` be the j`th matrix entry of B (relative to the

orthonormal bases of V2 and V1 in the statement of the theorem). Then the j`th entry of A (relative to the same bases) is equal to

n1 n2 −1 X X X 1 2 −1 |G| ajµ(g)bµν aν`(g ). g∈G µ=1 ν=1

Suppose that π1 6' π2. By the corollary to Schur’s Lemma, A = 0. Since this holds for all choices of B, we may choose B such that bµν = δµkδνm, 1 ≤ µ ≤ n1, 1 ≤ ν ≤ n2. Then −1 P 1 2 −1 2 |G| g∈G ajk(g)am`(g ) = 0. Since the matrix coeﬃcients am`(g) are chosen relative −1 2 to an orthonormal basis of V2 which makes π2 unitary, it follows that am`(g ) = a`m(g). 1 2 −1 P 1 2 Hence hajk, a`mi = |G| g∈G ajk(g)a`m(g) = 0. This proves (1). Now suppose that π1 = π2. In this case, Schur’s Lemma implies that A = λI for some −1 P −1 scalar λ. Hence tr A = |G| g∈G tr (π1(g)Bπ1(g) ) = tr B = n1λ. That is, the j`th entry of the matrix A is equal to n n −1 X X X 1 1 −1 |G| ajµ(g)bµν aν`(g ) = tr Bδj`/n1. g∈G µ=1 ν=1 −1 P 1 1 −1 Taking B so that bµν = δµkδνm, we have |G| g∈G ajk(g)am`(g ) = δj`δkm/n1. qed

15 Corollary. Let π1 and π2 be irreducible representations of G such that π1 6' π2. The susbpace of A(G) spanned by all matrix coeﬃcients of π1 is orthorgonal to the subspace spanned by all matrix coeﬃcients of π2. 1 Proof. Let ajk(g) be as in Theorem 3. Let γ be a basis of the space V1 of π1, and let bjk(g) be the jkth entry of the matrix [π1(g)]γ . Then there exists a matrix C ∈ GLn1 (C) such −1 that [bjk(g)] = C[ajk(g)]1≤j,k≤n1 C for all g ∈ G (C is the change of basis matrix from the β to γ). It follows that

1 bjk ∈ Span{ a`m | 1 ≤ m, ` ≤ n1 }.

Hence the subspace spanned by all matrix coefficients of π1 coincides with the subspace spanned by the matrix coefficients a`m, 1 ≤ `, m ≤ n1. Hence the corollary follows from Theorem 3(1). qed Corollary. There are finitely many equivalence classes of representations of a finite group G. Proof. This is an immediate consequence of the preceding corollary, together with dim A(G) = |G|. qed

For the remainder of this chapter, let G be a ﬁnite group, and let { π1, . . . , πr } be a complete set of irreducible representations of G, that is, a set of irreducible representations of G having the property that each irreducible representation of G is equivalent to exactly j one πj. Let nj be the degree of πj, 1 ≤ j ≤ r. Let a`m(g) be the `mth entry of the matrix of πj(g) relative to an orthonormal basis of the space of πj with respect to which each matrix of πj is unitary. √ j Theorem 4. The set { nja`m | 1 ≤ `, m ≤ nj, 1 ≤ j ≤ r } is an orthonormal basis of A(G). Proof. According to Theorem 3, the set is orthormal. Hence it suﬃces to prove that the set spans A(G). The regular representation RA is completely reducible. So A(G) = t ⊕k=1Vk, where each Vk is an irreducible G-invariant subspace. Fix k. There exists j such that RA |Vk ' πj. Choose an orthonormal basis β = { f1, . . . , fnj } of Vk such that j [RA(g) |Vk ]β = [a`m(g)], g ∈ G. Then

nj X j f`(g0) = (RA(g0)f`)(1) = ai`(g0)fi(1), 1 ≤ ` ≤ nj. i=1

Pnj j Hence f` = i=1 ciai`, with ci = fi(1). It follows that j Vk ⊂ Span{ a`m | 1 ≤ `, m ≤ nj }. qed

16 Theorem 5. Let 1 ≤ j ≤ r. The representation πj occurs as a subrepresentation of RA with multiplicity nj.

j j j Proof. Fix m ∈ { 1, . . . , nj }. Let Wm = Span{am` | 1 ≤ ` ≤ nj }. Then { am` | 1 ≤ ` ≤ nj } 0 j j j 0 0 is an orthogonal basis of Wm. And Wm is orthogonal to Wm0 whenever j 6= j or m 6= m . r nj j Hence A(G) = ⊕j=1 ⊕m=1 Wm. Let g, g0 ∈ G. Then

nj j j X j j RA(g0)am`(g) = am`(gg0) = amµ(g)aµ`(g0), 1 ≤ ` ≤ nj. µ=1

j j It follows that the matrix of RA(g0) relative to the basis { am` | 1 ≤ ` ≤ nj } of Wm nj j coincides with the matrix of πj. Therefore the restriction of RA to the subspace ⊕m=1Wm is equivalent to the nj-fold direct sum of πj. qed

2 2 Corollary. n1 + ··· + nr = |G|.

Corollary. A(G) equals the span of all matrix coeﬃcients of all irreducible representations of G.

Theorem 6 (Row orthogonality relations for irreducible characters). Let χj =

χπj , 1 ≤ j ≤ r. Then hχk, χj) = δjk.

Proof. n nk j X X j j 0, if k 6= j hχk, χji = haµµ, aνν i = Pnj . µ=1 1/nj = 1, if k = j µ=1 ν=1

Lemma. A ﬁnite-dimensional representation of a ﬁnite group is determined up to equivalence by its character.

Proof. If m is positive integer, let mπj = πj ⊕· · ·⊕πj, where πj occurs m times in the direct Pr 0 sum. Let π = m1π1 ⊕m2 ⊕· · ·⊕mrπr. Then χπ = j=1 mjχj. Let π = `1π1 ⊕· · ·⊕`rπr. 0 We know that π ' π if and only if mj = `j for 1 ≤ j ≤ r. By linear independence of the functions χj, this is equivalent to χπ = χπ0 . qed

Pr 2 Lemma. Let π = m1π1 ⊕ · · · ⊕ mrπr. Then hχπ, χπi = j=1 mj .

Corollary. π is irreducible if and only if hχπ, χπi = 1.

A complex-valued function on G is a class function if it is constant on conjugacy classes in G. Note that the space of class functions on G is a subspace of A(G) and the inner product on A(G) restricts to an inner product on the space of class functions.

17 Theorem 7. The set { χj | 1 ≤ j ≤ r } is an orthonormal basis of the space of class functions on G. Consequently the number r of equivalence classes of irreducible representations of G is equal to the number of conjugacy classes in G.

Proof. By Theorem 6, the set { χj | 1 ≤ j ≤ r } is orthonormal. It suﬃces to prove that the functions χj span the class functions. Let f be a class function on G. Since f ∈ A(G), we can apply Theorem 4 to conclude that r nj r nj X X √ √ X X f = hf, n aj i n aj = n hf, aj iaj . j m` j m` j m` m` j=1 m,`=1 j=1 m,`=1 Next,

r nj −1 X 0 0−1 −1 X X j X j 0 0−1 (∗) f(g) = |G| f(g gg ) = |G| nj hf, am`i am`(g gg ). g0∈G j=1 m,`=1 g0∈G

Note that

nj −1 X j 0 0−1 −1 X X j 0 j j 0−1 |G| am`(g gg ) = |G| amν (g )aµν (g)aν`(g ) g0∈G g0∈G µ,ν=1

nj   nj X −1 X j 0 j j X j j j = |G| amµ(g )a`ν (g) aµν (g) = hamµ, aν`iaµν (g) µ,ν=1 g0∈G µ,ν=1

nj −1 X j −1 = nj δm` aµµ(g) = δm`nj χj(g). µ=1

Substituting into (*) results in

r nj r nj ! r X X j X X j X f(g) = hf, am`iδm`χj(g) = hf, a``i χj(g) = hf, χjiχj(g). j=1 m,`=1 j=1 `=1 j=1 qed If g ∈ G, let |cl(g)| be the number of elements in the conjugacy class of g in G. Theorem 8 (Column orthogonality relations for characters).

r X |G|/|cl(g)|, if g0 is conjugate to g χ (g)χ (g0) = j j 0, otherwise. j=1

Proof. Let g1, . . . , gr be representatives for the the distinct conjugacy classes in G. Let

A = [χj(gk)]1≤j,k≤r. Let cj = |cl(gj)|, 1 ≤ j ≤ r. Let D be the diagonal matrix with

18 diagonal entries cj, 1 ≤ j ≤ r. Then

r r r ∗ X ∗ X X (ADA )m` = (AD)mjAj` = χm(gt)Dtjχ`(gj) j=1 j=1 t=1 r X X = χm(gj)cjχ`(gj) = χm(g)χ`(g) = |G|δm` j=1 g∈G

Thus ADA∗ = |G|I. Since A(DA∗) is a scalar matrix, A(DA∗) = (DA∗)A. So DA∗A = |G|I. That is, r r X ∗ X |G|δm` = (DA )mjAj` = cjχj(gm)χj(g`). j=1 j=1

qed

Example: Let G be a nonabelian group of order 8. Because G is nonabelian, we have Z(G) 6= G, where Z(G) is the centre of G. Because G is a 2-group, Z(G) 6= {1}. If |Z(G)| = 4, then |G/Z(G)| = 2, so G/Z(G) is cyclic. That is, G/Z(G) = hgZ(G)i, g ∈ G. Hence G = hZ(G)∪{x}i. But this implies that G is abelian, which is impossible. Therefore |Z(G)| = 2. Now |G/Z(G)| = 4 implies G/Z(G) is abelian. Combining G nonabelian and

G/Z(G) abelian, we get Gder ⊂ Z(G). We cannot have Gder trivial, since G is nonabelian.

So we have Gder = Z(G). Now, we saw above that G/Z(G) cannot be cyclic. Thus

G/Z(G) = G/Gder ' Z/2Z × Z/2Z.

Suppose that χ is a linear character of G (that is, a one-dimensional representation). −1 −1 −1 −1 Then χ | Gder ≡ 1, because χ(g1g2g1 g2 ) = χ(g1)χ(g2)χ(g1) χ(g2) = 1. Now Gder is a normal subgroup of G. So we can consider χ as a linear character of G/Gder. Now, in view of results on tensor products of representations, we know that G/Gder has 4 irreducible (one-dimensional) representations, each one being the tensor product of two characters of Z/2Z. Hence 2 2 2 2 2 2 1 + 1 + 1 + 1 + n5 + ··· + nr = |G| = 8, with nj ≥ 2, j ≥ 5. It follows that r = 5 and n5 = 2.

Since Gder = Z(G) has order 2, there are two conjugacy classes consisting of single elements. There are 5 conjugacy classes altogether. Let a, b, and c be the orders of the conjugacy classes containing more than 1 element. Then 2 + a + b + c = 8 implies a = b = c = 2. Let x1, x2 and x3 be representatives of the conjugacy classes containing 2 elements. Let z be the nontrivial element of Z(G). Then 1, y, x1, x2, x3 are representatives

19 of the 5 conjugacy classes. The character table of G takes the form: 1 1 2 2 2 1 y x1 x2 x3 χ1 1 1 1 1 1 χ2 1 1 1 −1 −1 χ3 1 1 −1 1 −1 χ4 1 1 −1 −1 1 χ5 2 χ5(y) χ5(x1) χ5(x2) χ5(x3) Using column orthogonality relations, we see that X 0 = χj(y)χj(1) = 4 + 2χ5(y), j=1 implying χ5(y) = −2. And

5 4 X X 0 = χj(xk)χj(1) = χj(xk) + 2χ5(xk) = 2χ5(xk), j=1 j=1 implying χ5(xk) = 0, 1 ≤ k ≤ 3. Note that (up to isomorphism) there are two nonabelian groups of order 8, the dihedral group D8, and the quaternion group Q8. We see from this example that both groups have the same character table. Exercises:

1. Using orthogonality relations, prove that if (πj,Vj) is an irreducible representation of

a ﬁnite group Gj, j = 1, 2, then π1 ⊗ π2 is an irreducible representation of G1 × G2. Then prove that every irreducible representation of G1 × G2 arises in this way.

2. Let D10 be the dihedral group of order 10.

a) Describe the conjugacy classes in D10. b) Compute the character table of D10.

3. Let B be the upper triangular Borel subgroup in GL3(Fp), where Fp is a ﬁnite ﬁeld containing p elements, p prime. Let N be the subgroup of B consisting of the upper triangular matrices having ones on the diagonal. a) Identify the set of one-dimensional representations of B. b) Suppose that π is an irreducible representation of B, having the property that

π(x)v 6= v for some x ∈ N and v ∈ V . Show that π |N is a reducible representation of N.(Hint: One approach is to start by considering the action of the centre of N on V .)

4. Suppose that G is a ﬁnite group. Let n ∈ N. Deﬁne θn : G → N by

n θn(g) = | { h ∈ G | h = g } |, g ∈ G.

20 Let χi, 1 ≤ i ≤ r be the distinct irreducible (complex) characters of G. Set

−1 X n νn(χi) = |G| χi(g ). g∈G P Prove that θn = 1≤i≤r νn(χi)χi. 5. Let (π, V ) be an irreducible representation of a ﬁnite group G. Prove Burnside’s Theorem:

Span{ π(g) | g ∈ G } = EndC(V ).

(Hint: Of course the theorem is equivalent to Span{ [π(g)]β | g ∈ G } = Mn×n(C), where β is a basis of V . This can be proved using properties of matrix coefficients of π (Theorem 3)). 6. Let (π, V ) be a finite-dimensional representation of a finite group G. Let

Wext = Span{ v ⊗ v | v ∈ V } ⊂ V ⊗ V . Wsym = Span{ v1 ⊗ v2 − v2 ⊗ v1 | v1, v2 ∈ V } ⊂ V ⊗ V

a) Prove that Wext and Wsym are G-invariant subspaces of V ⊗ V (considered as the space of the inner tensor product representation π ⊗ π of G). 2 2 2 b) Let (∧ π, ∧ V ) be the quotient representation, where ∧ V = (V ⊗ V )/Wext. 2 Then ∧ π is called the exterior square of π. Compute the character χ∧2π. c) Let (Sym2π, Sym2V ) be the quotient representation, where Sym2V = (V ⊗ 2 V )/Wsym. Then Sym π is called the symmetric square of π. Prove that (the inner tensor product) π ⊗ π is equivalent to ∧2π ⊕ Sym2π. 7. Let (π, V ) be the permutation representation associated to an action of a ﬁnite group

G on a set X. Show that χπ(g) is equal to the number of elements of X that are fixed by g. 8. Let f be a function from a finite group G to the complex numbers. For each finite dimensional representation (π, V ) of G, define a linear operator π(f): V → V by P π(f)v = g∈G f(g)π(g)v, v ∈ V . Prove that π(f) ∈ HomG(π, π) for all finite- dimensional representations (π, V ) of G if and only if f is a class function. 9. A (finite-dimensional) representation (π, V ) of a finite group is called faithful if the homomorphism π : G → GL(V ) is injective (one-to-one). Prove that every irreducible representation of G occurs as a subrepresentation of the set of representations

{ π, π ⊗ π, π ⊗ π ⊗ π, π ⊗ π ⊗ π ⊗ π, .... }.

10. Show that the character of any irreducible representation of dimension greater than 1 takes the value 0 on some conjugacy class.

21 11. A ﬁnite-dimensional representation (π, V ) of a ﬁnite group is multiplicity-free if each irreducible representation occurring in the decomposition of π into a direct sum of irreducibles occurs exactly once. Prove that π is multiplicity-free if and only if the

ring HomG(π, π) is commutative.

2.2. Character values as algebraic integers, degree of an irreducible represne- tation divides the order of the group

A complex number z is an algebraic integer if f(z) = 0 for some monic polynomial f having integer coeﬃcients. The proof of the following lemma is found in many standard references in algebra.

Lemma. (1) Let z ∈ C. The following are equivalent: (a) z is an algebraic integer (b) z is algebraic over Q and the minimal polynomial of z over Q has integer coeﬃ- cients. (c) The subring Z[z] of C generated by Z and z is a ﬁnitely generated Z-module. (2) The algebraic integers form a ring. The only rational numbers that are algebraic integers are the elements of Z.

Lemma. Let π be a ﬁnite-dimensional representation of G and let g ∈ G. Then χπ(g) is an algebraic integer.

Proof. Because G is ﬁnite, we must have gk = 1 for some positive integer k. Hence π(g)k = 1. It follows that every eigenvalue of π(g) is a kth root of unity. Clearly a kth

root of unity is an algebraic integer. Since χπ(g) is the sum of the eigenvalues of π(g), it follows from part (2) of the above lemma that χπ(g), being a sum of algebraic integers, is an algebraic integer. qed

Lemma. Let g1, . . . gr be representatives of the conjugacy classes in G. Let cj be the

number of elements in the conjugacy class of gj, 1 ≤ j ≤ r. Deﬁne fi ∈ A(G) by

fi(g) = cjχi(g)/χi(1), if g is conjugate to gj. Then fi(g) is an algebraic integer for 1 ≤ i ≤ r and all g ∈ G.

Proof. Let g0 ∈ G. As g ranges over the elements in the conjugacy class of gj, so does −1 g0gg0 . Therefore

  X X −1 X −1 πi(g) = πi(g0)πi(g)πi(g0) = πi(g0)  πi(g) πi(g0) .

g∈cl(gj ) g∈cl(gj ) g∈cl(gj )

22 So T := P π (g) belongs to Hom (π , π ). By irreducibility of π , T = zI for some g∈cl(gj ) i G i i i z ∈ C. Note that X tr T = χi(g) = cjχi(gj) = zχi(1),

g∈cl(gj ) so z = cjχi(gj)/χi(1). 0 Let g be an element of cl(gs). Let aijs be the number of ordered pairs (g , gˆ) such that 0 g gˆ = g. Note that aijs is independent of the choice of g ∈ cl(gs).

    X 0 X (ciχt(gi)/χt(1)) (cjχt(gj)/χt(1)) I =  πt(g )  πt(ˆg) 0 g ∈cl(gi) gˆ∈cl(gj ) r X X 0 X X = πt(g gˆ) = aijsπt(g) 0 g ∈cl(gi) gˆ∈cl(gj ) s=1 g∈cl(gs) r ! X = aijscsχt(gs)/χt(1) I s=1 Hence r X (ciχt(gi)/χt(1)) (cjχt(gj)/χt(1)) = aijscsχt(gs)/χt(1). s=1

This implies that the subring of C generated by the scalars csχt(gs)/χt(1), 1 ≤ s ≤ r, and Z is a finitely-generated Z-module. Since Z is a principal ideal domain, any submodule of a finitely-generated Z-module is also a finitely-generated Z-module, the submodule Z[ciχt(gi)/χt(1)] is finitely-generated. Applying part (2) of one of the above lemmas, the result of this lemma follows. qed

Theorem 9. nj divides |G|, 1 ≤ j ≤ r. Proof. Note that

|G|/χi(1) = |G|hχi, χii/χi(1) r r X X = cjχi(gj)χi(gj)/χi(1) = (cjχi(gj)/χi(1)) χi(gj). j=1 j=1

Because the right side above is an algebraic integer, the left side is a rational number which is also an algebraic integer, hence it is an integer. qed

2.3. Decomposition of ﬁnite-dimensional representations In this section we describe how to decompose a representation π into a direct sum of j irreducible representations, assuming that the functions am` are known.

23 Lemma. Let (π, V ) be a ﬁnite-dimensional representation of G. For 1 ≤ k ≤ r, 1 ≤ j, ` ≤ k nk, deﬁne Pj` : V → V by

k −1 X k Pj` = nk|G| aj`(g)π(g). g∈G

Then k Pnk k k k Pnk k (1) π(g)Pj` = ν=1 aνj(g)Pν` and Pj`π(g) = ν=1 a`ν (g)Pjν , g ∈ G. k k0 k 0 (2) Pj`Pµν = Pjν if k = k and ` = µ, and equals 0 otherwise. k ∗ k (3) (Pj`) = P`j. Proof. For the ﬁrst part of (1),

k −1 X k 0 0 −1 X k −1 0 0 π(g)Pj` = |G| nk aj`(g )π(gg ) = |G| nk aj`(g g )π(g ) g0∈G g0∈G

nk nk −1 X X k −1 k 0 0 X k k = |G| nk ajν (g )aν`(g )π(g ) = aνj(g)Pν` g0∈G ν=1 ν=1

The second part of (1) is proved similarly. For (2),

nk k k0 −1 X k k0 −1 X k X k0 k0 Pj`Pµν = nk|G| aj`(g)π(g)Pµν = nk|G| aj`(g) atµ(g)Ptν g∈G g∈G t=1

nk X k0 k k0 k = nk hatµ, aj`iPtν = δkk0 δ`µPjν . t=1

For (3),

k ∗ −1 X k −1 −1 X k −1 −1 k (Pj`) = nk|G| aj`(g)π(g ) = nk|G| a`j(g )π(g ) = P`j. g∈G g∈G

k k k ∗ k k 2 k Set Vj = Pjj(V ). Note that , since (Pjj) = Pjj = (Pjj) , Pjj is the orthogonal k k k0 projection of V on Vj . From property (2) of the above lemma, it follows that Vj ⊥ Vj0 if k 6= k0 or j 6= j0. r nk k k k k k ⊥ Let W = ⊕k=1 ⊕j=1 Vj . Note that Pj`(V ) = PjjPj`(V ) ⊂ Vj ⊂ W . Fix v0 ∈ W . Then k −1 X k k 0 = hv0,Pj`(v)i = nk|G| aj`(g)hv0, π(g)viV = nkhaj`, fiA(G), g∈G

⊥ where f(g) = hv0, π(g)vi, g ∈ G. It follows that f ∈ A(G) . Hence f(g) = 0 for all g ∈ G.

Setting v = v0 and g = 1, we have hv0, v0iV = 0. Thus v0 = 0. That is, W = V .

24 k k0 0 Next, note that Pj`Vt = 0 if k 6= k or t 6= `, by part (2) of the above lemma. k k 0 0 k 0 k k 0 Let v ∈ V` . Then v = P``(v ) for some v ∈ V . Now v = P``(v ) = P``(P``(v )) = k P``(v), so we have k k k k k k Pj`(v) = Pj`P``(v) = PjjPj`(v) ⊂ Vj .

k k k Thus Pj`(V` ) ⊂ Vj . Now

k k k k k k k k k V` = P``V` = P`jPj`V` ⊂ P`jVj ⊂ V` .

k k k 0 k Hence we have P`jVj = V` . Let v, v ∈ V` . Then

k k 0 k ∗ k 0 k k 0 k 0 0 hPj`(v),Pj`(v )i = h(Pj`) Pj`(v), v i = hP`jPj`(v), v i = hP`,`(v), v i = hv, v i.

We have shown

k k k Lemma. Pj` is an isometry of V` onto Vj . Choose an orthonormal basis ek , ek , . . . , ek of V k = P k (V ). Then ek := P k (ek ), 11 21 rk,1 1 11 j` `1 j1 k 1 ≤ j ≤ rk is an orthonormal basis of V` . It follows that the set

k { ej` | 1 ≤ j ≤ rk, 1 ≤ ` ≤ nk, 1 ≤ k ≤ r }

k k is an orthonormal basis of V . Set Yj = Span{ ej` | 1 ≤ ` ≤ nk }. If g ∈ G, then

nk nk k k k X k k k X k k π(g)ej` = π(g)P`1(ej1) = aν`(g)Pν1ej1 = aν`(g)ejν . ν=1 ν=1

k k This shows that Yj is G-invariant and has the matrix [aν`(g)]{1≤ν,`≤nk} relative to the k given orthonormal basis of Y . This implies that π | k ' πk, 1 ≤ j ≤ rk. Now j Yj k V = ⊕1≤k≤r ⊕1≤j≤rk Yj , so we have decomposed π into a direct sum of irreducible representations. This decomposition is not unique. k rk k k k Set Y = ⊕j=1Yj . Now { ej` | 1 ≤ j ≤ rk, 1 ≤ ` ≤ nk } is an orthonormal basis of Y , k k k nk k and π |Y k ' rkπk. Because P`` is the orthogonal projection of V on V` and Y = ⊕`=1V` , k Pnk k k it follows that P := `=1 P`` is the orthogonal projection of V on Y . Looking at the deﬁnitions, we see that this orthogonal projection P k is deﬁned by

nk k X −1 X k −1 X P = nk|G| a``(g)π(g) = nk|G| χk(g)π(g). `=1 g∈G g∈G

Suppose that W is a G-invariant subspace of V such that π |W is equivalent to a direct sum of πk with itself some number of times. Let { v1, . . . , vnk } be an orthonormal basis of

25 an irreducible G-invariant subspace of W , chosen so that the matrix of the restriction of k π(g) to this subspace is [a`m(g)]. Then

nk nk k −1 X −1 X X k X k P (vj) = nk|G| χk(g)π(g)vj = nk|G| a``(g) aµj(g)vµ g∈G g∈G `=1 µ=1

nk nk −1 X X k k X k k = nk|G| a``(g)aµj(g)vµ = nk haµj, a``ivµ g∈G `,µ=1 `,µ=1 −1 = nknk vj = vj Therefore P k | W is the identity. Because P k is the orthogonal projection of V on Y k, we know that P k(v) = v if and only if v ∈ Y k. It follows that W ⊂ Y k. Now we may conclude that if we have a G-invariant subspace of V such that the restriction of π to that subspace k is equivalent to rkπk, then that subspace must equal Y . Lemma. The subspaces Y k are unique.

k The subspace Y is called the πk-isotypic subspace of V . It is the (unique) largest subspace of V on which the restriction of π is a direct sum of representations equivalent k to πk. Of course, we will have Y = {0} if no irreducible constituent of π is equivalent to

πk.

2.4. Induced representations One method of producing representations of a finite group G is the process of induction: given a representation of a subgroup of G, we can define a related representation of G. Let (π, V ) be a (finite-dimensional) representation of a subgroup H of G. Define

V = { f : G → V | f(hg) = π(h)f(g), h ∈ H, g ∈ G }.

G G G We deﬁne the induced representation iH π = IndH (π) by (iH π(g)f)(g0) = f(g0g), g, g0 ∈ G. Observe that if h ∈ H, then

G G (iH π(g)f)(hg0) = f(hg0g) = π(h)f(g0g) = π(h)(iH π(g)f)(g0).

G −1 It follows from the deﬁnitions that the degree of iH π equals |G||H| times the degree of −1 P π. Let h·, ·iV be any inner product on V . Set hf1, f2iV = |G| g∈Ghf1(g), f2(g)iV , f1, f2 ∈ V. It is easy to check that this deﬁnes an inner product on V with respect to which G iH π is unitary. G Example If H = {1} and π is the trivial representation of H, then iH π is the right regular representation of G. G The Frobenius character formula expresses the character of iH π in terms of the character of π.

26 Theorem 10 (Frobenius character formula). Let (π, V ) be a representation of a

subgroup H of G. Fix g ∈ G. Let h1, . . . , hm be representatives for the conjugacy classes in H which lie inside the conjugacy class of g in G. Then

m −1 X −1 χ G (g) = |G||H| |clH (hi)||clG(g)| χπ(hi). iH π i=1

P G 0 Proof. Let g ∈ G. Deﬁne T : V → V by T = g0∈cl(g) iH π(g ). Note that tr T = |cl(g)|χ G (g). iH π Let β = { v1, . . . , vn } be an orthonormal basis of V such that each [π(h)]β, h ∈ H, is a unitary matrix. For each j ∈ {1, . . . , n}, deﬁne |G|1/2|H|−1/2π(h)v , if g = h ∈ H f (g) = j j 0, if g∈ / H.

1/2 −1/2 Then fj(h0g) = |G| |H| π(h0h)v = π(h0)fj(h)v, if g = h ∈ H, and fj(h0g) = fj(g) = 0 if g∈ / H. Thus fj ∈ V. Note that

−1 X −1 X hfj, fkiV = |G| hfj(g), fk(g)iV = |H| hπ(h)vj, π(h)vkiV g∈G h∈H −1 X = |H| hvj, vkiV = hvj, vkiV = δjk. h∈H

Therefore { f1, . . . , fn } is an orthonormal set in V. Pick representatives g1, . . . , g` of the cosets in H\G (that is, of the right H cosets in G). Then

G G −1 X −1 X −1 hiH π(gi)fj, iH π(gk))fsiV = |G| hfj(ggi), fs(ggk)iV = |G| hfj(h), fs(hgi gk)iV g∈G h∈H −1 X = δik|G| hπ(h)fj(1), π(h)fs(1)iV δikhfj, fsiV = δikδjs h∈H

G −1 Hence { iH π(gi)fj }1≤i≤`, 1≤j≤n is an orthonormal set. Since dim V = |G||H| dim V = G `n, it is an orthonormal basis of V (with respect to which iH π is a unitary representation). The kkth entry of T with respect to this basis is

hT (kth basis element), kth basis elementiV .

Therefore ` n X X G G tr T = hT (iH π(gi)fj), iH π(gi)fjiV . i=1 j=1

0 0 −1 As g ranges over the conjugacy class cl(g), gig gi also ranges over cl(g). Hence

G G −1 (iH π)(gi)T iH π(gi ) = T,

27 and the above expression for tr T becomes

` n n X X G G X h(iH π)(gi)T fj, (iH π)(g0)fjiV = ` hT fj, fjiV . i=1 j=1 j=1

Now we rewrite each hT fj, fjiV using the deﬁnitions of fj and T .

X −1 X 0 X −1 X 0 hT fj, fjiV = |G| hfj(g0g ), fj(g0)iV = |G| hfj(hg ), fj(h)iV 0 0 g ∈cl(g) g0∈G g ∈cl(g)∩H h∈H m X −1 X = |clH (hs)||G| hfj(hhs), fj(h)iV s=1 h∈H m m −1 X X X = |H| |clH (hs)| hπ(h)π(hs)vj, π(h)vjiV = |clH (hs)|hπ(hs)vj, vjiV s=1 h∈H s=1

It follows that

m n m −1 X X −1 X tr T = |G||H| |clH (hs)| hπ(hs)vj, vjiV = |G||H| |clH (hs)|χπ(hs). s=1 j=1 s=1

−1 Pm −1 Thus χ G (g) = |G||H| |clG(g)| |clH (hs)|χπ(hs). qed iH π s=1 Example: Applying the Frobenius character formula with π the trivial representation of the trivial subgroup of G, we see that the character of the regular representation of G vanishes on all elements except for the identity element.

The inner product on A(G) restricts to an inner product on the space C(G) of class functions on G. When we wish to identify the fact that we are taking the inner product on G A(G), we will sometimes write h·, ·iG. Let H be a subgroup of G. We may view iH as a map G from C(H) to C(G), mapping χπ to i (χπ) := χ G , for π any irreducible representation H iH π of H. As the characters of the irreducible representations of H form a basis of C(H), the H map extends by linearity to all of C(H). We can deﬁne a linear map rG from C(G) to C(H) by restricting a class function on G to H. The next result, Frobenius Reciprocity, H G tells us that rG is the adjoint of the map iH . If (π, V ) is a representation of G and τ is an irreducible representation of G, the multiplicity of τ in π is deﬁned to be the number of times that τ occurs in the decomposition of π as a direct sum of irreducible representations of G. This multiplicity is equal to hχτ , χπiG = hχπ, χτ iG. Theorem 11 (Frobenius Reciprocity). Let (π, V ) be an irreducible representation of H H and let (τ, W ) be an irreducible representation of G. Then hχτ , χ G iG = hr χτ , χπiH . iH π G

28 Proof. Let π and τ be as in the statement of the theorem. Let g ∈ G be such that

clG(g)∩H 6= ∅. Choose h1, . . . , hm as in the previous theorem. Then, using χτ (g) = χτ (hi), 1 ≤ i ≤ m, m ! −1 X −1 χ G (g)χτ (g) = |G||H| |clH (hi)||clG(hi)| χπ(hi) χτ (g) iH π i=1 m −1 X −1 = |G||H| |clH (hi)||clG(hi)| χπ(hi)χτ (hi) i=1 −1 P Now when evaluating hχ G , χτ iG = |G| χ G (g)χτ (g), we need only sum over iH π g∈G iH π those g ∈ G such that cl(g) ∩ H 6= ∅. Then

−1 X hχ G , χτ iG = |H| χπ(h)χτ (h) = hχπ, rGH (χτ )iH . iH π h∈H qed Corollary (Transitivity of induction). Suppose that K ⊂ H are subgroups of G. Let G G H (π, V ) be a representation of K. Then iK π = iH (iK π). K K H Proof. Note that it follows from the definitions that rG = rH ◦ rG . Taking adjoints, we have G K ∗ H ∗ K ∗ G H iK = (rG ) = (rG ) ◦ (rH ) = iH ◦ iK . qed Lemma. Let (π, V ) be a representation of a subgroup H of G. Fix g ∈ G. Let π0 be the −1 0 −1 G G 0 representation of gHg defined by π (ghg ) = π(h), h ∈ H. Then iH π ' igHg−1 π . G −1 Proof. Let f be in the space of iH π. Set (Af)(g0) = f(g g0), g0 ∈ G. Let h ∈ H. Then −1 −1 0 −1 (Af)(ghg g0) = f(hg g0) = π(h)(Af)(g0) = π (ghg )(Af)(g0). G 0 Therefore Af belongs to the space of igHg−1 π . It is clear that A is invertible. Note that −1 G igHg−1 (g1)(Af)(g0) = (Af)(g0g1) = f(g g0g1) = (A iH π(g1)f)(g0). qed Let (π, V ) be a finite-dimensional representation of a subgroup H of G. Let K be a subgroup of G, and let g ∈ G. Then K ∩gHg−1 is a subgroup of G. Define a representation πg of this subgroup by πg(k) = π(g−1kg), k ∈ K ∩ gHg−1. Let h ∈ H. Then K ∩ ghH(gh)−1 = K ∩ gHg−1 and πgh(k) = π(h−1g−1kgh) = π(h)−1πg(k)π(h). Hence πgh ' g K gh K g π . This certainly implies that iK∩ghHhg−1 π ' iK∩gHg−1 π . K g Changing notation slightly, we see that the above lemma tells us that iK∩gHg−1 π ' kg K g iK∩kgHg−1k−1 π . We now know that the equivalence class of iK∩gHg−1 π is independent of the choice of element g inside its K-H-double coset (that is, we may replace g by kgh, k ∈ K, h ∈ H, without changing the equivalence class).

29 Theorem 12. (Mackey) K and H be subgroups of G, and let (π, V ) be a representation of H. Then G K G M K g (iH π)K = rG (iH π) ' iK∩gHg−1 (π ). g∈K\G/H G Proof. Let ρ = iH π. Let V be the space of ρ. Deﬁne a map A : V → V by (Av)(g) = π(h)v, if g = h ∈ H , v ∈ V . Let v ∈ V , and g , g ∈ G. Then ρ(g )Av ∈ ρ(g )AV 0, if g∈ / H. 1 1 2 1 1 2 −1 −1 if and only if ρ(g2 g1)Av1 = Av2 for some v2 ∈ V . Now ρ(g2 g1)Av1 is supported in −1 Hg1 g2 and Av2 is supported in H. Hence the two functions are equal if and only if P g1H = g2H. It follows that g∈G/H ρ(g)AV = ⊕g∈G/H ρ(g)AV . Now ρ(g) is invertible and A is one-to-one, so dim ρ(g)AV = dim V . Therefore the dimension of the latter direct −1 sum equals |G||H| dim V = dim V. Thus V = ⊕g∈G/H ρ(g)AV . Now we want to study V as a K-space and ρ(g)AV is not K-stable. Given g ∈ G, the double coset KgH is a disjoint union of certain cosets g0H. So we group together 0 0 P 0 those ρ(g )AV such that g H ⊂ KgH. Let X(g) = g0H⊂KgH ρ(g )AV . It should be understood that the above sum is taken over a set of representatives g0 of the left H-cosets which lie in KgH. Now we have regrouped things and we have V = ⊕g∈K\G/H X(g). Now ρ(k)X(g) = X(g) for all k ∈ K. We will prove that

K K g (∗∗) ρK |X(g) = rG ρ |X(g) ' iK∩gHg−1 π . The theorem is a consequence of (**) and the above direct sum decomposition of V. Now suppose that g0 = kgh, k ∈ K, h ∈ H. Then ρ(g0)AV = ρ(kg)AV . Let −1 k0 ∈ K ∩ gHg .

0 −1 ρ(k0g )AV = ρ(k0g)AV = ρ(g(g k0g))AV = ρ(g)AV P . This implies that X(g) = K/(K∩gHg−1) ρ(k)ρ(g)AV . Now we can easily check that if k ∈ K, then ρ(k)ρ(g)AV = ρ(g)AV if and only if ρ(g−1kg)AV = AV if and only if −1 −1 g kg ∈ H, that is k ∈ K ∩ gHg . So X(g) = ⊕K/(K∩gHg−1) ρ(k)ρ(g)AV . K g Let W be the space of iK∩gHg−1 π . Now define B : X(G) → W as follows. Let v ∈ V g −1 and k ∈ K. Set ϕv(k) = π (k)v if k ∈ K ∩gHg and ϕv(k) = 0 otherwise. Then ϕv ∈ W. K g Given v ∈ V and k ∈ K, set Bρ(k)ρ(g)Av = iK∩gHg−1 π (k)ϕv. It is a simple matter to K K g check that B is invertible. Since rG ρ acts by right translation on X(g) and iK∩gHg−1 π acts by right translation on W, we see that B intertwines these representations. Hence (**) holds. qed Theorem 13. Let H and K be subgroups of a finite group G. Let (π, V ) and (ρ, W ) be G G (finite-dimensional) representations of H and K, respectively. Then HomG(iH π, iK ρ) is isomorphic to

{ ϕ : G → EndC(V,W ) |ϕ(kgh) = ρ(k) ◦ ϕ(g) ◦ π(h), k ∈ K, g ∈ G, h ∈ H }.

30 G G Sketch of proof. Let A ∈ HomG(iH π, iK ρ). Deﬁne ϕA : G → EndC(V,W ) by ϕA(g)v = (Afv)(g), v ∈ V , where fv(h) = π(h)v, h ∈ H, and fv(g) = 0 if g∈ / H. Note that fv is in G the space of iH π. Let k ∈ K, g ∈ G, h ∈ H, and v ∈ V . Then

ϕA(kgh)v = (Afv)(kgh) = ρ(k)(Afv)(gh) = ρ(k)(ρ(h)Afv)(g)

= ρ(k)(Afπ(h)v)(g) = ρ(k)ϕA(g)π(h)v

G where the second equality holds because Afv is in the space of iK ρ, the fourth equality G G holds because A ∈ HomG(iH π, iK ρ), and the ﬁfth because π(h)fv = fπ(h)v for all h ∈ H. Hence ϕA has the desired properties relative to left translation by elements of K and right translation by elements of H.

Given ϕ : G → EndC(V,W ) | ϕ(kgh) = ρ(k)◦ϕ(g)◦π(h), g ∈ G, h ∈ H, k ∈ K. Let f be in the space of iG π. Deﬁne A f in the space of iG ρ by (A f)(g) = P ϕ(gg−1)f(g ), H ϕ K ϕ g0 0 0 where in the sum g0 runs over a set of coset representatives for K\G. Suppose that k ∈ K and g ∈ G. Then

X −1 X −1 (Aϕf)(kg) = ϕ(kgg0 )f(g0) = ρ(k) ϕ(gg0 )f(g0) = ρ(k)(Aϕf)(g). g0 g0

G Hence Aϕf belongs to the space of iK ρ. Next, let g, g1 ∈ G. Then

G X −1 G X −1 (AϕiH π(g1)f)(g) = ϕ(gg0 )(iH π(g1)f)(g0) = ϕ(gg0 )f(g0g1) g0 g0 X −1 G = ϕ(gg1g0 )f(g0) = (Aϕf)(gg1) = (iK ρ(g1)Aϕf)(g) g0

G G Therefore Aϕ ∈ HomG(iH π, iK ρ). To ﬁnish the proof, check that A 7→ ϕA and ϕ 7→ Aϕ are inverses of each other. The details are left as an exercise. qed

G G Corollary. Let π be an irreducible representation of a subroup H of G. Then HomG(iH π, iH π) is isomorphic to

0 0 0 H(G, π) := { ϕ : G → EndC(V ) | ϕ(hgh ) = π(h) ◦ ϕ(g) ◦ π(h ), g ∈ G, h, h ∈ H }. Lemma. The subspace of H(G, π) consisting of functions supported on the double coset g Hg g −1 g −1 HgH is isomorphic to HomHg (π , rH π). where H = H ∩ gHg and π (h) = π(g hg), h ∈ Hg. Proof. Fix g ∈ G. Given ϕ ∈ H(G, π) such that ϕ is supported on HgH, deﬁne a linear g operator Bϕ : V → V by Bϕ(v) = ϕ(g)v, v ∈ V . Then, if h ∈ H , we have, using the fact that g−1hg ∈ H and h ∈ H, and properties of ϕ,

g −1 Bϕ(π (h)v) = ϕ(g)(π(g hg)v) = ϕ(hg)v = π(h)ϕ(g)v = π(h)Bϕ(v).

31 g Hg Hence Bϕ ∈ HomHg (π , rH π). g Hg Given B ∈ HomHg (π , rH π), set ϕB(h1gh2)v = π(h1)Bπ(h2)v, for h1, h2 ∈ H and v ∈ V , and ϕB(g1)v = 0 if g1 ∈/ HgH. Check that ϕB ∈ H(G, π), and also that the map

B 7→ ϕB is the inverse of the map ϕ 7→ Bϕ. The details are left as an exercise. qed Corollary. Let (π, V ) be a representation of a subgroup H of G. If g ∈ G, let Hg = H ∩ gHg−1 and set πg(h) = π(g−1hg), g ∈ Hg. Then

G G M g Hg HomG(iH π, iH π) ' HomHg (π , rH π). g∈H\G/H

Corollary(Mackey irreducibility criterion). Let (π, V ) be an irreducible representa- G g Hg tion of a subgroup H of G. Then iH π is irreducible if and only if HomHg (π , rH π) = 0 for all g∈ / H.

h Hh Proof. Note that if g = h ∈ H, then H = H and rH π = π. Hence, by irreducibility of π, h G G G HomH (π, π ) ' C. Therefore, since iH π is irreducible if and only if HomG(iH π, iH π) ' C, G g Hg by the above proposition, iH π is irreducible if and only if HomHg (π , rH π) = 0 whenever g∈ / H. qed

G G Corollary. If π is the trivial representation of a subgroup H of G, then dim HomG(iH π, iH π) equals the number of H-H-double cosets in G.

g Hg g Note that π and rH π are both the trivial representation of H (for any g ∈ G). Acccord- G G ing to the above proposition, there is a one-dimensional contribution to HomG(iH π, iH π) for each double coset HgH. qed

Given ϕ1, ϕ2 ∈ H(G, π), set

−1 X −1 (ϕ1 ∗ ϕ2)(g) = |G| ϕ1(g0) ◦ ϕ2(g0 g). g0∈G

This product makes H(G, π) into an algebra, known as a Hecke algebra.

G G Proposition. The algebra HomG(iH π, iH π) is isomorphic to the Hecke algebra H(G, π).

To prove the proposition, check that the vector space A 7→ ϕA of the the Theorem 13 is an algebra homomorphism: ϕA1◦A2 = ϕA1 ∗ ϕA2 . The details are left as an exercise. Exercises: 1. Adapt the above arguments to prove: Let H and K be subgroups of G, let π be a G G representation of H, and let τ be a representation of K. Suppose that iH π and iK τ G G are irreducible. Show that iH π 6' iK τ if and only if for every g ∈ G

g gHg−1∩K HomgHg−1∩K (π , rK τ) = 0.

32 2. Let A be an abelian subgroup of a group G. Show that each irreducible representation of G has degree at most |G||A|−1.

3. Let πj be a representation of a subgroup Hj of a group Gj, j = 1, 2. Prove that

iG1×G2 π ⊗ π ' iG1 π ⊗ iG2 π . H1×H2 1 2 H1 1 H2 2

4. Let π and ρ be representations of subgroups H and K of a ﬁnite group G. Let

g1 g2 −1 −1 g1, g2 ∈ G. Deﬁne representations π and ρ of the group g1 Hg1 ∩ g2 Kg2 by g1 −1 g2 −1 −1 −1 π (x) = π(g1xg1 ) and ρ (x) = ρ(g2xg2 ), x ∈ g1 Hg1 ∩ g2Kg2 . (g1,g2) G g1 g2 (i) Prove that the equivalence class of τ := i −1 −1 (π ⊗ ρ ) depends g1 Hg1∩g2Kg2 −1 only on the double coset Hg1g2 K. G G (ii) Prove that the internal tensor product iH π ⊗ iK ρ is equivalent to the direct sum (g1,g2) −1 of the representations τ as g1g2 ranges over a set of representatives for the H-K-double cosets in G.(Hint: Consider the restriction of the representation G G ¯ of G × G which is the outer tensor product iH π ⊗ iK ρ to the subgroup G = { (g, g) | g ∈ G } and apply Theorem 11.) 5. Suppose that the ﬁnite group G is a the semidirect product of a subgroup H with an abelian normal subgroup A, that is, G = H n A. Let G act on the set Aˆ of irreducible (that is, one-dimensional) representations of A by σ 7→ σg, where σg(a) = σ(g−1ag),

a ∈ A, σ ∈ Aˆ. Let { σ1, . . . , σr } be a set of representatives for the orbits of G on Aˆ. h Let Hi = { σi = σi }. a) Let Gi = Hi n A. Show that σi extends to a representation of Gi via σi(ha) = σi(a), h ∈ Hi, a ∈ A.

b) Let π be an irreducible representation of Hi. Show that we may deﬁne an irreducible representation ρ(π, σi) of Gi on the space V of π by: ρ(π, σi)(ha) =

σi(ha)π(h) = σi(a)π(h), h ∈ Hi, a ∈ A. c) Let π be an irreducible representation of H . Set θ = IndG ρ(π, σ ). Prove that i i,π Gi i θi,π is an irreducible representation of G. 0 d) Let π and π be irreducible representations of Hi and Hi0 . Prove that θi,π ' θi0,π0 implies i = i0 and π ' π0.

e) Prove that { θi,π } are all of the (equivalence classes of) irreducible representations of G. (Here, i ranges over {1, . . . , r} and, for i ﬁxed, π ranges over all of the

(equivalence classes of) irreducible representations of Hi). 6. Let H be a subgroup of a ﬁnite group G. Let H(G, 1) be the Hecke algebra associated with the trivial representation of the subgroup H. a) Show that if (π, V ) is an irreducible representation of G, and V H is the subspace of H-ﬁxed vectors in V , then V H becomes a representation of H(G, 1), that is, a

33 module over the ring H(G, 1), with the action

X f · v = |G|−1 f(g)π(g)v, v ∈ V H . g∈G

b) Show that if V H 6= {0}, then V H is an irreducible representation of H(G, 1) (an irreducible H(G, 1)-module). (Hint: If W is a nonzero invariant subspace of V H , H and v ∈ V , use irreducibility of π to show that there exists a function f1 on G

such that v = f1 · w, where w ∈ W and f1 · w is deﬁned as above even though

f1 ∈/ H(G, 1). Next, show that if 1H is the characteristic function of H, then

f := 1H ∗ f1 ∗ 1H , f ∈ H(G, 1), and f · w = v.) c) Show that (π, V ) → V H is a bijection between the equivalence classes of irreducible representatations of G such that V H 6= {0} and the equivalence classes of irreducible representations of H(G, 1).

Remarks - Representations of Hecke algebras: Suppose that (π, V ) is an irreducible finite-dimensional representation of a subgroup H of G. A representation of the Hecke 0 algebra H(G, π) is defined to be an algebra homomorphism from H(G, π) to EndC(V ) for some finite-dimensional complex vector space V 0. (That is, V 0 is a finite-dimensional H(G, π)-module). Let (ρ, W ) be a finite-dimensional representation of G. Then it is easy to check that H the internal tensor product (rG ρ ⊗ π, W ⊗ V ) contains the trivial representation of H if H ∨ and only if rG ρ contains the representation π of H that is dual to π. ∨ H Assume that the dual representaion π is a subrepresentation of rG ρ. Given f ∈ H(G, π), define a linear operator ρ0(f) on W ⊗ V by

X ρ0(f)(w ⊗ v) = |G|−1 ρ(g)w ⊗ f(g)v. g∈G

The fact that f ∈ H(G, π) can be used to prove that the subspace (W ⊗ V )H of H- invariant vectors in W ⊗ V is ρ0(f)-invariant, and the map f 7→ ρ0(f) | (W ⊗ V )H deﬁnes a representation of the Hecke algebra H(G, π). In this way, we obtain a map ρ 7→ ρ0 from the set of representations of G whose restrictions to H contain π∨ and the set of nonzero representations of the Hecke algebra H(G, π). It can be shown that this map has the following properties: (i) ρ is irreducible if and only if ρ0 is irreducible. 0 0 (ii) If ρ1 and ρ2 are irreducible, then ρ1 and ρ2 are equivalent if and only if ρ1 and ρ2 are equivalent. (iii) For each nonzero irreducible representation (τ, U) of H(G, π), there exists an irreducible representation ρ of G such that ρ0 is equivalent to τ.

34 The study of representations of reductive groups over finite fields (that is, finite groups of Lie type) is sometimes approached via the study of representations of Hecke algebras. In certain cases, H(G, π) may be isomorphic (as an algebra) to another Hecke algebra H(G0, π0), where G0 is a different group (and π0 is an irreducible representation of a subgroup H0 of G0). In this case, the study of those irreducible representations of G whose restrictions to H contain π∨ reduces to the study of a similar set of representations of the group G0. Representations of Hecke algebras also play a role in the study of admissible representations of reductive groups over p-adic fields. An example of such a group is GLn(Qp) where Qp is the field of p-adic numbers. In this setting, the representation ρ of G will be infinite-dimensional (and admissible), the subgroup H will be compact, and open in G, π will be finite-dimensional (since H is compact) and the representation ρ0 of the Hecke algebra will be finite-dimensional. In this setting, the definition of the Hecke algebras is slightly different from that for finite groups.

35 CHAPTER 3

Representations of SL2(Fq)

Let Fq be a ﬁnite ﬁeld of order q. Then there exists a prime p and a positive integer ` such that q = p`. For convenience, we assume that p is odd. We also assume that q 6= 3. Let G = SL2(Fq) be the group of 2 × 2 matrices with entries in Fq and determinant equal to 1. In this chapter, we construct the (characters of the) irreducible representations of G. a b Let B = | a ∈ ×, b ∈ . Let A be the group of diagonal matrices 0 a−1 Fq Fq 1 x in G, and let N = | x ∈ . Note that B = A N ' × . Hence 0 1 Fq n Fq n Fq |B| = q(q − 1). a b Let g ∈ G. Note that g = ∈/ B if and only if c 6= 0. In that case, b = c d −1 × c (ad−1). Because a and d may be chosen freely in Fq and c may be chosen freely in Fq , it follows that the number of elements in the complement of B in G is equal to q2(q − 1). Hence |G| = q(q − 1) + q2(q − 1) = q(q2 − 1). 0 1 Let w = . The double coset BwB is a disjoint union of right B cosets. Let −1 0 −1 −1 b1, b2 ∈ B. Then Bwb1 = Bwb2 if and only if b1b2 ∈ w Bw ∩ B. Note that

a b a−1 0 (3.1) w−1 w = . 0 a−1 −b a

It follows that wBw−1 ∩ B = A, and as g ranges over a set of (right) coset representatives for A\B, then g ranges over a set of representatives for the right B cosets in BwB. Note that A\B ' N. Hence we may (and do) view N as a set of coset representatives for A\B and for the right B cosets in BwB. Now |N| = q. Thus BwB contains exactly q right B cosets. Hence |BwB| = q|B| = q2(q − 1). The subset B q BwB contains q(q − 1) + q2(q − 1) = q(q2 − 1) elements, so must equal G. Lemma. G is generated by B and w, and there are two B-B double cosets in G, namely B and BwB. Also 1 x G = q Bw q B. x∈Fq 0 1 We will begin to study the representations of G by looking at representations of G which are induced from linear characters (one-dimensional representations) of B. Note that the derived group (commutator subgroup) of B is equal to N. Hence there is a bijection between the set of linear characters of A ' B/N and the set of linear characters × × of B. Now A ' Fq and Fq is a cyclic group of order q −1. Let ζ ∈ C be a primitive root of × unity of order q − 1, and let α be a generator of Fq . For each m such that 0 ≤ m ≤ q − 2,

36 j jm × the map α 7→ ζ deﬁnes a linear character of Fq . It is clear that these characters are distinct. Since there are q − 1 of them, this gives a complete list of the linear characters × × of Fq . Note that there are two characters of Fq whose squares are trivial. One is the × trivial representation of Fq , corresponding to m = 0, and the other one, corresponds to × m = (q − 1)/2, and takes the value −1 on non-squares in Fq and 1 on squares. a b Let τ be a character of ×. The associated character τ of B is deﬁned by τ = 0 Fq 0 a−1 w w −1 τ0(a). Looking at equation (3.1), we see that the character τ of B = wBw ∩ B = A (notation as in Chapter 2) is given by (3.2) a 0 a 0 a−1 0 τ w = τ w−1 w−1 w = τ = τ (a)−1, a ∈ ×. 0 a−1 0 a−1 0 a 0 Fq Hence 2 w Bw −1 C, if τ0 = 1 HomBw (τ , rB τ) = HomA(τ0 , τ0) = 2 0, if τ0 6= 1 Combining this with results from Chapter 2 concerning induced representations, we have

G 2 2 G G Lemma. iBτ is irreducible if and only if τ0 6= 1. If τ0 = 1, then dim HomG(iBτ, iBτ) = 2. a 0 Given a ∈ ×, set s = . It is easy to check that if a 6= ±1, then A is the Fq a 0 a−1 centralizer of sa in G. Hence if a 6= ±1, the order |cl(sa)| of the conjugacy class cl(sa) in −1 G is equal to |G|/|A| = q(q + 1). Note that if a, b ∈ Fq, then gsag = sb if and only if sa and sb have the same eigenvalues. If a 6= ±1, this is equivalent to b ∈ { sa, sa−1 }. −1 Thus, when a 6= ±1, cl(sa) = cl(sb) if and only if b = a or b = a . Now the centre of G

is equal to { s1, s−1 }, so there are two single-element conjugacy classes in G. According to the above there are (q − 3)/2 non-central conjugacy classes which contain elements of × A, each such class containing q(q + 1) elements. If a ∈ Fq and a 6= ±1, then it is easy to −1 check that clB(sa) = { gsag | g ∈ N } = saN. We can now conclude that

× clG(sa) ∩ B = saN q sa−1 N, a ∈ Fq , a 6= ±1.

Note that clG(±I) = ±I = clG(±I) ∩ B. Applying the Frobenius Character Formula, we can compute χ G on A. iB τ × Lemma. Let a ∈ Fq . Then

−1 −1 −1 χ G (sa) = |G||B| |clG(sa)| (|clB(sa)|τ(sa)+|clB(sa−1 )|τ(sa−1 )) = τ0(a)+τ0(a ), a 6= ±1. iB τ and χ G (s±1) = (q + 1)τ0(±1). iB τ

Let ψ be a nontrivial character of (the additive group) Fq. Let t ∈ Fq. It is easy to see that the function x 7→ ψt(x) := ψ(tx) also deﬁnes a character of Fq, and ψs = ψt if

37 and only if s = t. Hence { ψt | t ∈ Fq } is the set of characters of Fq. We can also view this set as the set of characters of N, since N ' Fq. τ τ Let t ∈ Fq. Deﬁne a function ft : G → C by ft (b) = 0 for b ∈ B and

1 x f τ bw = τ(b)ψ (x), b ∈ B, x ∈ . t 0 1 t Fq

τ τ G o Clearly ft 6= 0 and ft belongs to the space of iBτ. Let Vτ be the subspace of the space G Vτ of iBτ which consists of those functions which are supported in BwB. Then the set τ o { ft | t ∈ Fq } is a basis of Vτ . Note that

1 y 1 x 1 x + y iGτ f τ bw = f τ bw B 0 1 t 0 1 t 0 1 1 x = τ(b)ψ (x + y) = ψ (y)f τ bw . t t t 0 1

τ N G Therefore (for t ﬁxed) the span of ft is N-invariant and the restriction of rG (iBτ) to this one-dimensional space equivalent to the character ψt of N. It follows that the N G o restriction of rG (iBτ) to Vτ is equivalent to the regular representation of N, since it is the direct sum of all of the characters of N, each occurring exactly once. τ τ Let fB be the function which is zero on BwB and satisﬁes fB(b) = τ(b), b ∈ B. The 0 space Vτ of Vτ consisting of functions supported on B is one-dimensional, and is spanned τ N G 0 by the function fB. It is clear that the restriction of rG (iBτ) to Vτ is equivalent to the trivial representation of N. Lemma. N G L (1) r (i τ) ' 2ψ0 ⊕ × ψt. G B t∈Fq 1 x q + 1, if x = 0 (2) χiGτ = . B 0 1 1, if x 6= 0 × Lemma. If g ∈ G has no eigenvalues in , then χ G (g) = 0. Fq iB τ

Proof. If χ G (g) 6= 0, then by the Frobenius Character Formula, clG(g) ∩ B 6= ∅. Any iB τ × element of B has eigenvalues in Fq . qed 0 × 0 Lemma. Let τ0 and τ be characters of , with extensions τ and τ to B. Then χ G = 0 Fq iB τ 0 ±1 χ G 0 if and only if τ0 = (τ ) . iB τ 0 −1 0 −1 Proof. Note that τ0(−1) = τ0 (−1). Hence τ0(−1) = τ0(−1) if and only if (τ0+τ0 )(−1) = 0 0−1 (τ + τ )(−1). Given the above results about χ G and χ G 0 , we see that χ G = χ G 0 0 0 iB τ iB τ iB τ iB τ if and only if −1 0 0−1 × (τ0 + τ0 )(a) = (τ0 + τ0 )(a), ∀ a ∈ Fq . × 0±1 By linear independence of characters of Fq , this is equivalent to τ0 = τ0 . qed

38 0 2 Corollary. Let τ and τ0 be as above Assume that τ0 6= 1. Then

, if τ 0 = τ ±1 Hom (iGτ 0, iGτ) = C 0 0 G B B 0, otherwise.

0 2 G G 0 Proof. Suppose that (τ0) = 1. Then iBτ is irreducible of degree q+1 and iBτ is reducible of degree q + 1. Therefore the two representations do not have any irreducible constituents in common. 0 2 G G 0 Now suppose that (τ0) 6= 1 Then iBτ and iBτ are irreducible and by the previous 0±1 result are equivalent if and only if τ0 = τ0 . qed From above, we can conclude that there are have (q − 3)/2 equivalence classes of G 2 irreducible representations of the form iBτ (of course with τ 6= 1). G 2 The next step is to decompose iBτ in the two cases with τ = 1. Lemma. The trivial representation of G occurs as a subrepresentation of the representa- G tion iB1 induced from the trivial representation of B. The other irreducible constituent of G i 1 has degree q and its character is equal to χ G minus the characteristic function of G. B iB 1 G Proof. We already know that iB1 has two irreducible constituents and has degree q + 1. Therefore it suﬃces to prove that the trivial representation of G occurs as a subrepresen- G G tation of iB1. The characteristic function of G belongs to the space of iB1 and it is clearly invariant under right translation by elements of G. (Although it is not needed for the proof G of the lemma, it is worth noting that the space of the other irreducible constituent of iB1 P consists of functions which are left B-invariant and which satisfy f(1)+ u∈N f(wu) = 0). qed × × 2 Note that, since we have assumed that p is odd, Fq /(Fq ) has order 2. Fix a nonsquare × × × 2 ε ∈ Fq . Then 1 and ε are coset representatives for Fq /(Fq ) . Let λ be the unique character × τ τ of Fq of order 2. Let ft = ft (for τ corresponding to λ), t ∈ Fq, and let fB = fB. Let G ρ = iBτ. We know that G is generated by N, A and w. Hence a subspace of the space Vλ is G-invariant if and only if it is N, A, and w-invariant. We have already described the N-invariant subspaces. Next, we describe some B-invariant subspaces. Let

× 2 × 2 W1 = Span{ ft | t ∈ (Fq ) } and Wε = Span{ ft | t ∈ ε(Fq ) }.

1 x If x ∈ , let u = . Fq x 0 1

× Lemma. Let a ∈ Fq . × (1) If t ∈ Fq , then ρ(sa)ft = λ(a)fa−2t. (2) ρ(sa)f0 = λ(a)f0

(3) ρ(sa)fB = λ(a)fB.

39 (4) W1, Wε, Span{f0}, and Span{fB} are B-invariant. × Proof. Part (4) follows from parts (1)–(3). For parts (1) and (2), let a ∈ Fq , t ∈ Fq and b ∈ B. Then −1 (ρ(sa)ft)(bwux) = ft(bwuxsa) = ft(bsa−1 wua−2x) = λ(a )τ(b)ft(wua−2x) −2 = λ(a)τ(b)ψt(a x) = λ(a)τ(b)ψa−2t(x) = λ(a)fa−2t(bwux). −1 −1 Above we have used the fact that bsa ∈ sa−1 bN and τ is trivial on N, and λ = λ . Part (3) is left as an exercise. qed Exercise: Prove that there are 4 inequivalent irreducible representations of B that are not one-dimensional, and they all have degree (q − 1)/2. Now we consider the action of w. It is easy to see that, as representations of B,

W1 and Wε are irreducible. It follows from results above that they are inequivalent, since as representations of N they have no constituents in common. As we see below, no subspace of Span{ fB, f0 } is w-invariant. Hence W1 and Wε cannot belong to the same G G-subrepresentation of iBτ. We will use × (3.2) wuxw = s−x−1 wu−x−1 , x ∈ Fq G Let ρ = iBτ, where τ corresponds to λ.

Lemma. Let t ∈ Fq. −1 × (1) Then (ρ(w)ft)(bwux) = ft(bwuxw) = λ(−x)ft(bwu−x−1 ) = λ(−x)ψt(−x ), x ∈ Fq , b ∈ B,

(2) (ρ(w)ft)(bw) = 0, b ∈ B. (3) (ρ(w)ft)(b) = τ(b), b ∈ B. Let X hϕ1, ϕ2i = ϕ1(1)ϕ2(1) + ϕ1(wux)ϕ2(wux), ϕ1, ϕ2 ∈ Vτ . x∈Fq Lemma.

(1) h·, ·i is an inner product on Vτ with respect to which ρ is a unitary representation of G. (2) { fB, ft | t ∈ Fq } is an orthogonal basis of Vτ (relative to the given inner product). Proof. Part (1) is easily veriﬁed using the decomposition of G given in the ﬁrst lemma. For part (2), note that if t ∈ Fq, the support of ft does not intersect B. Hence hft, fBi = 0. Let s, t ∈ Fq. Then X hft, fsi = ψt(x)ψs(x) = q δst, x∈Fq using orthogonality relations of characters of Fq. qed P Let Γ = × λ(x)ψ(x). x∈Fq

40 Lemma. −1 P (1) ρ(w)fB = q λ(−1) ft. t∈Fq −1 P (2) ρ(w)f0 = fB + q Γ × λ(t)ft. t∈Fq Proof. For (1), note that

−1 X −1 X 2 q λ(−1) ft(w) = q λ(−1) ψt(0) = λ(−1) = fB(w ) = (ρ(w)fB)(w) t∈Fq t∈Fq

× and, for x ∈ Fq ,

−1 X −1 X −1 X q λ(−1) ft(wux) = q λ(−1) ψt(x) = q λ(−1) ψx(t) = 0 = (ρ(w)fB)(wux), t∈Fq t∈Fq t∈Fq

−1 × since ψx is a nontrivial character of Fq, and wuxw ∈/ B if x ∈ Fq . P From the previous lemma, we have ρ(w)f0 = fB + ctft for some scalars ct. And t∈Fq because { fB, ft | t ∈ Fq } is an orthogonal basis of Vτ , we have ct = hρ(w)f0, fti/hft, fti = −1 q hρ(w)f0, fti, t ∈ Fq. Because ft(1) = (ρ(w)f0)(w) = f0(−1) = 0, we have

X X −1 X hρ(w)f0, fti = ρ(w)f0(wux)ft(wux) = λ(−x)ψ0(−x )ψt(x) = λ(−x)ψt(−x) × × × x∈Fq x∈Fq x∈Fq ( P −1 P × x∈ × λ(t x)ψ(x) = λ(t) x∈ × λ(x)ψ(x) = λ(t)Γ, if t ∈ Fq = P Fq Fq × λ(x) = 0, if t = 0 x∈Fq Lemma. Γ2 = qλ(−1).

P × Proof. Let Φ(t) = × ψ(tx)λ(x), t ∈ . Then x∈Fq Fq X Φ(t) = ψ(y)λ(t)−1λ(y) = λ(t)Γ. × y∈Fq

× Then, if x ∈ Fq , X X Φ(t)ψ(tx) = Γ λ(t)ψ(tx) = ΓΦ(x) = Γ2λ(x). × × t∈Fq t∈Fq But we also have   X X X X X Φ(t)ψ(tx) = ψ(tu + tx)λ(u) =  ψ(t(u + x)) λ(u) × × × × × t∈Fq t∈Fq u∈Fq u∈Fq t∈Fq . X = λ(−x)(q − 1) − λ(u) = λ(−x)(q − 1) − (0 − λ(−x)) = λ(−x)q × u∈Fq , u6=−x Therefore Γ2λ(x) = λ(−x)q. This implies Γ2 = λ(−1)q. qed

41 Lemma. −1 P (1) ρ(w)(Γf0 ± qfB) = ±q Γ(Γf0 ± qfB) + λ(−1) × (λ(t) ± 1)ft. t∈Fq (2) ρ(w)(Γf0 + qfB) ∈ Span(Γf0 + qfB) + W1. (3) ρ(w)(Γf0 − qfB) ∈ Span(Γf0 − qfB) + Wε.

+ − Set V = Span(Γf0 + qfB) + W1 and V = Span(Γf0 − qfB) + Wε. We can see that V+ = (V−)⊥ and V− = (V+)⊥. The above lemma suggests that V+ and V− may be + − G-invariant. We need to show that ρ(w)W1 ⊂ V and ρ(w)Wε ⊂ V . Lemma. × × 2 (1) Let s, t ∈ Fq . Then hρ(w)ft, fsi = 0 if s∈ / t(Fq ) . × 2 (2) If t ∈ (Fq ) , then hρ(w)ft, Γf0 − qfBi = 0. × 2 (3) If t ∈ ε(Fq ) , then hρ(w)ft, Γf0 + qfBi = 0. × Proof. Let s, t ∈ Fq . Then

X X −1 X −1 hρ(w)ft, fsi = ρ(w)ft(wux)fs(wux) = λ(x)ψt(x )ψs(x) = λ(x)ψ(tx +sx). × × × x∈Fq x∈Fq x∈Fq

2 × Now suppose that s = εtu for some u ∈ Fq . Then

X −1 X −1 2 X −1 2 λ(x)ψ(t x + sx) = ψt(x + xεu ) − ψt(εx + xu ). × × 2 × 2 x∈Fq x∈(Fq ) x∈(Fq )

2 −1 2 × 2 −1 2 The map x 7→ xu is a bijection from { x + xεu | x ∈ (Fq ) } and { εx + xu | x ∈ × 2 × (Fq ) }. It follows that hρ(w)ft, fεtu2 i = 0 for all u ∈ Fq . × 2 For (2), let t ∈ (Fq ) . Then

hρ(w)ft, Γf0 − qfBi = hft, ρ(−w)(Γf0 − qfB)i −1 X = λ(−1)hft, −Γq (Γf0 − qfB) + λ(−1) −2fsi = 0. × 2 s∈ε(Fq )

The proof of part (3) is similar to that of part (2) and is left as an exercise. qed Corollary. V+ and V− are G-invariant.

+ − Let πλ = ρ |V+ and πλ = ρ |V− . For the proof of the following lemma, see the discussion on conjugacy classes which appears on pages 31.

× Lemma. Let x ∈ Fq . 2 (1) |clG(ux)| = (q − 1)/2. × 2 (2) clG(ux) = clG(u1) if x ∈ (Fq ) .

42 × 2 (3) clG(ux) = clG(uε) if x ∈ ε(Fq ) . (4) clG(u1) 6= clG(uε).

+ − + − Theorem. ρ = πλ ⊕ πλ , πλ and πλ are irreducible and inequivalent. Furthermore, (1) χ + (s±1) = λ(−1)(q + 1)/2. πλ × 2 (2) χ + (sa) = λ(a), a ∈ Fq , a 6= 1. πλ (3) χ + (±u1) = λ(−1)(1 + Γ)/2. πλ (4) χ + (±uε) = λ(−1)(1 − Γ)/2. πλ × (5) χ + (g) = 0 if g has no eigenvalues in Fq . πλ (6) χ − = χρ − χ + = χiGτ − χ + (where τ is the character of B corresponding to λ). πλ πλ B πλ + − Proof. Because the restrictions of πλ and πλ to N are inequivalent, the representations are inequivalent. For part (1), note that dim V+ = (q + 1)/2 and ρ(−1) = λ(−1)I. × 2 −2 For part (2), recall that ρ(sa)ft = λ(a)fa t, t ∈ (Fq ) . Therefore tr (ρ(sa) |W1 ) = 0 2 × if a 6= 1. Also, ρ(sa)(Γf0 + qfB) = λ(a)(Γf0 + qfB), a ∈ Fq . Let x ∈ Fq. Set X X X X h1(x) = ψt(x) = ψx(t) and hε(x) = ψt(x) = ψx(t). × 2 × 2 × 2 × 2 t∈(Fq ) t∈(Fq ) t∈ε(Fq ) t∈ε(Fq )

× If x ∈ Fq , then h1(x) + hε(x) = −1 because ψx is a nontrivial character of Fq. Also, if × x ∈ Fq , X X −1 h1(x) − hε(x) = λ(t)ψx(t) = λ(x s)ψ(s) = λ(x)Γ. × × t∈Fq s∈Fq

Solving for h1(x) and hε(x), we get h1(x) = (−1 + λ(x)Γ)/2 and hε(x) = (−1 − λ(x)Γ)/2, × x ∈ Fq . Now tr (ρ(ux) |W1 ) = h1(x) and ρ(ux)(Γf0 + qfB) = Γf0 + qfB. Therefore χ + (u1) = h1(1) + 1 = (1 + Γ)/2. Similarly, χ + (uε) = h1(ε) + 1 = (1 − Γ)/2. Parts (3) πλ πλ and (4) follow. × Let Gspl be the set of elements of G which have eigenvalues in Fq . Any element in × Gspl is conjugate to one of sa, a ∈ Fq or one of ±u1 and ±uε. Therefore

X 2 2 2 |χ + (g)| = 2((q + 1)/2) + q(q + 1)(q − 3)/2 + 2((1 + Γ + Γ¯ + ΓΓ)¯ /4)(q − 1)/2 πλ g∈Gspl + 2((1 − Γ − Γ¯ + ΓΓ)¯ /4)(q2 − 1)/2 = (q + 1)2/2 + q(q + 1)(q − 3)/2 + (1 + ΓΓ)(¯ q2 − 1)/2

Now X X Γ¯ = λ(x)ψ(−x) = λ(−1) λ(x)ψ(x) = λ(−1)Γ. × × x∈Fq x∈Fq

43 2 2 P 2 Hence ΓΓ¯ = λ(−1)Γ = λ(−1) q = q. Substituting above results in |χ + (g)| = g∈Gspl πλ 2 + P 2 q(q − 1) = |G|. Because π is irreducible, we have |χ + (g)| = |G|. Therefore λ g∈G πλ χ + (g) = 0 if g∈ / Gspl qed πλ Before moving on to ﬁnding the other irreducible characters of G, we discuss conjugacy classes further. Let g ∈ G be unipotent (that is, (g − 1)2 = 0). Then both eigenvalues

of g equal 1. Suppose that g 6= 1 and g is unipotent. Then, because u1 is the Jordan −1 canonical form of g, there exists g0 ∈ GL2(Fq) such that g0gg0 = u1. That is, there is one noncentral unipotent conjugacy class in GL2(Fq). This class breaks up into several 2 conjugacy classes in G. The centralizer of u1 in G is ±N, so |clG(u1)| = (q − 1)/2. × From sauxsa = ua2x, x ∈ Fq , we see that ua2 ∈ clG(u1). Let d(ε) be the diagonal matrix in GL2(Fq) having diagonal entries ε and 1, respectively. If uε ∈ clG(u1), then, −1 −1 −1 choosing g ∈ G such that gu1g = iε, we have d(ε)u1d(ε) = gu1g . This implies a b g−1d(ε) centralizes u . Therefore g−1d(ε) = for some a ∈ F × and some b ∈ . 1 0 a q Fq Taking determinants, we have ε = a2, which is impossible, since ε is a non-square. Hence clG(u1) 6= clG(uε).

Now suppose that g ∈ Gell. Then the eigenvalues of g, being roots of the characteristic polynomial of g are roots of a quadratic polynomial which is irreducible over Fq. Hence the eigenvalues lie in a quadratic extension of . Up to isomorphism, there is exactly one Fq √ quadratic extension of . Since ε is a non-square in ×, ( ε) is a quadratic extension Fq √ Fq Fq of . Hence the eigenvalues of g lie in ( ε) − . Now g ∈ G, so the product of the Fq Fq Fq √ eigenvalues equals 1. It follows that the eigenvalues of g are of the form a + b ε and √ × 2 2 a − b ε where a ∈ Fq, b ∈ Fq , and a − b ε = 1. An example of such a g is the matrix a bε , with b 6= 0 and a2 − b2ε = 1. Let b a

a bε T = | a, b ∈ , a2 − b2ε = 1 . b a Fq

Then T is a subgroup of G and any element of T −{±I} belongs to G . Note that { 1, ε } is √ √ ell a basis of Fq( ε), and the matrix of an elment a+b ε with respect to this basis is equal to a bε √ . The map N : ( ε)× → × is a surjective homomorphism. Hence the kernel b a Fq Fq √ a bε √ of N has order (q2 − 1)/(q − 1) = q + 1. The map a + b ε 7→ from ( ε)× to b a Fq GL2(Fq) restricts to an isomorphism between N and T . Hence |T | = q+1. Also, T is cyclic, √ × × since it is a subgroup of the cyclic group Fq( ε) ' Fq2 . It is a simple matter to check that if γ ∈ T −{±I}, then the centralizer of γ in G is equal to T . Hence |clG(γ)| = q(q−1). a bε Now let γ = ∈ T be such that γ2 6= 1. Any element of T which is conjugate b a

44 √ √ √ to γ must have the same eigenvalues as γ, namely a + b ε and a − b ε = (a + b ε)−1. −1 −1 Checking that γ and γ are conjugate in G, we have clG(γ) ∩ T = {γ, γ }. There are q − 1 noncentral elements in T . Thus there are exactly (q − 1)/2 noncentral conjugacy classes which intersect T , each containing q(q − 1) elements. Proposition. (1) The centre of G equals {±I}. × (2) If a ∈ Fq and a 6= ±1, then |clG(sa)| = q(q+1). There are exactly (q−3)/2 noncentral conjugacy classes which intersect A. 2 (3) cl(u1) 6= cl(uε), cl(−u1) 6= cl(−uε), and |cl(±uε)| = |cl(±u1)| = (q − 1)/2. (4) If γ ∈ T and γ 6= ±1, then |cl(γ)| = q(q − 1). There are exactly (q − 1)/2 noncentral conjugacy classes which intersect T . (5) There are q + 4 conjugacy classes in G. Proof. The only part which has not already been proved is part (5). First, counting the conjugacy classes mentioned in parts (1)–(4), we have a total of

2 + (q − 3)/2 + 4 + (q − 1)/2 = q + 4.

Counting the number of elements in the unions of all of these conjugacy classes, we get

2 + q(q + 1)(q − 3)/2 + 4(q2 − 1)/2 + q(q − 1)(q − 1)/2 = q(q2 − 1) = |G|.

The number of distinct irreducible characters of G constructed so far (from the rep- G resentations iBτ and their irreducible constituents) is equal to (q − 3)/2 + 4 = (q + 5)/2, (q − 3)/2 of degree q + 1, one of degree 1, one of degree q, and two of degree (q + 1)/2. We must ﬁnd another (q + 3)/2 irreducible characters. We have found the characters of those irreducible representations of G whose restrictions to N contain the trivial representations of N (equivalently whose restrictions to B contain a one-dimensional representation of B). An irreducible representation of G is said to be cuspidal if its restriction to N does not contain the trivial representation of N (equivalently, its restriction to B contains no one-dimensional representations of B). As we will see, the nontrivial characters of the group T will determine the other irreducible characters of G, but not in exactly the same way that the characters of A determined the characters already constructed.

Exercise: For each x ∈ Fq, deﬁne

+ − ψ (±ux) = ψ(x), ψ (±ux) = ±ψ(x) + − ψε (±ux) = ψε(x), ψε (±ux) = ±ψε(x).

± B ± ± B ± Let σ1 = i±N ψ , and σε = i±N ψε . Prove that

45 ± ± (1) σ1 and σε are inequivalent irreducible representations of B, and any irreducible representation of B of degree greater than 1 is equivalent to one of them.

(2) χ ± (±u1) = χ ± (±uε) = ±(Γ − 1)/2 σ1 σε (3) χ ± (±uε) = χ ± (±u1) = ±(−Γ − 1)/2. σ1 σε (4) All four of the above characters vanish on B − ±N.

A class function ϕ on a ﬁnite group G0 is a virtual character of G0 if ϕ is an integral linear combination of characters of irreducible representations of G. Because the sum and product of the characters of representations of G0 are also characters of representations (the product being the character of a tensor product), the set of virtual characters of G0 is a ring

(with pointwise addition and multiplication of functions). Let χ1, . . . , χr be the characters 0 Pr of a complete set of irreducible representations of G . Suppose that ϕ = j=1 `jχj. Then Pr 2 hϕ, ϕiA(G0) = j=1 `j . It follows that if hϕ, ϕi = 1, then ϕ = ±χj for some j. If, in addition, ϕ(1) > 0, then ϕ = χj. Note that we cannot take this approach with arbitrary Pr 2 class functions on G because we can ﬁnd complex numbers c1, . . . , cr with j=1 |cj| = 1 Pr without forcing j=1 cjχj to be a multiple of one χj. One approach to determining irreducible characters of a ﬁnite group G0 is to generate as many virtual characters ϕ as possible, and then look for those which satisfy hϕ, ϕiA(G) = 1. Let π be an irreducible cuspidal representation of G. Because the restriction of π to the subgroup B is the direct sum of irreducible representations of B of degree greater than

1, and every irreducible representation of B has degree (q − 1)/2, χπ(1) is divisible by G ± G ± (q − 1)/2. By Frobenius reciprocity, π occurs as a subrepresentatation of iBσ1 = i±N ψ G ± G ± or iBσε = i±N ψε . The proof of the following lemma is left as an exercise.

Lemma. 2 (1) χiG ψ± (±1) = χ G ± (±1) = ±(q − 1)/2. ±N i±N ψε (2) χ G ± (±u1) = ±(Γ − 1)/2 i±N ψ (3) χ G ± (±uε) = ±(−Γ − 1)/2 i±N ψ (4) χ G ± (±u1) = ±(−Γ − 1)/2 i±N ψε (5) χ G ± (±uε) = ±(Γ − 1)/2 i±N ψε 2 2 (6) χ G ± and χiG ψ± vanish on elements sa ∈ A with a 6= 1 and γ ∈ T with γ 6= 1. i±N ψε ±N

The irreducible characters constructed so far as all constant on Gell. Therefore the restrictions of the characters of the cuspidal representations of G must separate the conjugacy classes cl(γ), γ ∈ T such that γ2 6= 1. From the above lemma, we see that the characters χiG ψ± and χ G ± vanish on Gell, so the characters of the elliptic reprsenta- ±N i±N ψε tions cannot all be expressed as linear combinations of these characters. The nontrivial characters of T do separate these conjugacy classes in Gell. So perhaps the characters of G the representations iT θ for θ ranging over nontrivial characters of T will be related to the

46 characters of the cuspidal representations of G. The proof of the following lemma is left as an exercise.

Lemma. Let θ be a character of T . Then

(1) χ G (±1) = θ(±1)q(q − 1) iT θ −1 2 (2) χ G (γ) = θ(γ) + θ(γ ) if γ ∈ T and γ 6= 1. iT θ × 2 (3) Let a ∈ be such that a 6= 1. Then χ G (sa) = χ G (±u1) = χ G (±uε) = 0. Fq iT θ iT iT

Since the functions χiG ψ± , χ G ± and χiGθ are virtual characters of G, so is any ±N i±N ψε T integral linear combination of these functions. We will look for virtual characters ϕ of this form which satisfy hϕ, ϕi = 1. Suppose that π is an irreducible representation of G. Since −I belongs to the centre of G, it follows from Schur’s Lemma that π(−I) is a scalar multiple of the identity operator. And (−I)2 = I forces the scalar multiple to equal ±1. Looking back at the irreducible G representations of the form iBτ, on the unipotent set, the character formula involves the trivial character of N (appearing twice) and sum χiG ψ+ + χ G + or χiG ψ− + χ G − , ±N i±N ψε ±N i±N ψε with + if the above scalar is −1 and − if the scalar is −1 (note that this scalar equals τ(−1)). On noncentral elements of of A, the sum τ + τ −1 appears. With this in mind, knowing that the trivial representation of N will not occur in the character of a cuspidal representation, suppose that θ is a character of T , and set

( `χiGθ + m(χiG ψ+ + χ G + , if θ(−1) = 1 T ±N i±N ψε ϕθ = `χiGθ + m(χiG ψ− + χ G − , if θ(−1) = −1. T ±N i±N ψε where m, ` ∈ {±1}. Now if ϕθ is the character of a cuspidal representation, we must have 2 ϕθ(1) = `q(q − 1) + m(q − 1) = (q − 1)(`q + m(q + 1)) > 0, equal to a multiple of (q − 1)/2, and dividing q(q2 − 1) = |G|. Checking the possibilities, we must have m = 1 and ` = −1, so ϕθ(±1) = θ(−1)(q − 1). × The values of ϕθ on ±uε, ±u1, γ ∈ T and sa, a ∈ Fq may be obtained from character values given in the previous two lemmas.

Lemma.

(1) ϕθ(±1) = θ(−1)(q − 1)

(2) ϕθ(±u1) = ϕθ(±uε) = −θ(±1). 2 −1 (3) If γ ∈ T and γ 6= 1, then ϕθ(γ) = −θ(γ) − θ(γ ). × 2 (4) If a ∈ Fq and a 6= 1, then ϕθ(sa) = 0. Lemma. 1, if θ2 6= 1 hϕ , ϕ i = . θ θ A(G) 2, if θ2 = 1

47 2 2 Proof. According to the above lemma, |ϕθ(±u1)| = |ϕθ(±uε)| = 1. Thus

2 X q(q − 1)hϕθ, ϕθiA(G) = ϕθ(g)ϕθ(g) g∈G X = 2(q − 1)2 + 4(q2 − 1)/2 + q(q − 1)(1/2) (θ(γ) + θ(γ−1))2 γ∈T, γ6=±1 X = 4q(q − 1) + q(q − 1)/2 (θ(γ)2 + 2 + θ(γ−1)2) γ∈T γ26=±1   X 2 = 4q(q − 1) + q(q − 1) ( θ(γ) ) − 2 + (q − 1) γ∈T q − 3, if θ2 6= 1 = 4q(q − 1) + q(q − 1) 2(q − 1), if θ2 = 1. q(q2 − 1), if θ2 6= 1 = 2q(q2 − 1), if θ2 = 1. Theorem. Let θ be a character of T . 2 (1) If θ 6= 1, there exists an irreducible representation πθ of G such that χπθ = ϕθ. Also, If 0 0 ±1 θ is a character of T with nontrivial square, then πθ ' πθ0 if and only if θ = θ . Also

πθ is not equivalent to any of the irreducible representations obtained as irreducible constituents of representations induced from one-dimensional representations of B. (2) Let θ be a character of T such that θ2 = 1. Then there exist irreducible representations + − π and π of G such that ϕθ = χ + ± χ − . θ θ πθ πθ Proof. Apply the previous result, together with comments on properties of virtual charac- 2 ters, to obtain the proofs of the assertions regarding irreducibility of πθ when θ 6= 1.

From the values of χπθ = ϕθ, we can see that χπθ = χπθ0 if and only if the functions θ + θ−1 and θ0 + (θ0)−1 agree on T . Hence, by linear independence of characters of T , 0 ±1 we have πθ ' πθ0 if and only if θ = θ . Clearly the character πθ is distinct from the G characters of any of the irreducible constituents of the representations iBτ. 2 For (2), assume that θ = 1. We know that ϕθ is a virtual character. It follows from hϕθ, ϕθi = 2 that 2 equals the sums of squares of the integers occuring as coefficients in the expression of ϕθ as a linear combination of irreducible characters. Hence exactly 2 of the coefficients are nonzero, each one in the set {±1}. Now ϕθ(1) > 0. So it is clear that + − πθ and πθ can be chosen so that the signs must be as stated, qed At this point, ignoring the representations in part (2) above (whose characters we have G not yet computed), we have produced (q − 1)/2 irreducible characters of the form iBτ, 2 G irreducible characters of constitutents of iB1, χ + and χ − , and (q − 1)/2 inequivalent πλ πλ irreducible characters χπθ . That is, we have produced q + 2 distinct irreducible characters of G. There are (q + 4) − (q + 2) = 2 remaining to find (and both are cuspidal).

48 Computing the sums of the squares of the degrees of the irreducible characters already 2 produced, we obtain |G|−(q−1) /2. Therefore, if d1 and d2 are the degrees of the remaining 2 2 2 irreducible characters, we have d1 + d2 = (q − 1) /2. We already know that d1 and d2 are divisible by (q − 1)/2 (the degree of any irreducible representation of B that is not

one-dimensional). Therefore d1 = d2 = (q − 1)/2. 2 + − Suppose that θ is a character of T such that θ = 1. Let πθ and πθ be as above. Suppose that one of them is not cuspidal (that is, one of them contains a one-dimensional B representation of B). Then, because rG ϕθ is a combination of the characters of B of degree (q − 1)/2, the other one must also be non-cuspidal, and ϕθ = χ + − χ − . Furthermore, πθ πθ + − at least one of πθ and πθ must have a character which is nonvanishing on Gell, because ϕθ is not nonzero on Gell. As we have seen, there are exactly two equivalence classes of non-cuspidal irreducible representations of G whose characters are not identically zero on G Gell. They are the two constitutents of iB1. One is the trivial representation of G, and the other has degree q. Every other non-cuspidal irreducible representation has degree q + 1. + − Now ϕθ(1) = χ + (1) = χ − (1) = q−1. It follows that π has degree q and π is the trivial πθ πθ θ θ representation of G. Looking at the values of the characters of these representations, and of ϕθ, we see that we must have θ trivial.

Lemma. Let θ be the trivial character of T . Then ϕθ is the diﬀerence of the irreducible character of degree q and the trivial representation of G. Now let ν be the unique character of T of order 2. So ν takes the value −1 on non- squares in T , and 1 on squares in T . According to the comments above, we know that + − both πν and πν are cuspidal. Now they must have degree a multiple of (q − 1)/2. From the form of ϕν , we see that it may be the case that ϕν = χ + − χ − where χ + (1) = q − 1 πν πν πν and χ − (1) = (q −1)/2. In that case, since we know that the only cuspidal representations πν 2 which are not equivalent some πθ with θ 6= 1 all have degree (q − 1)/2, it follows that + 2 2 πν ' πθ for some θ with θ 6= 1. But we can check that hϕν , ϕθiA(G) = 0 if θ 6= 1. Therefore it is not possible for χ + to have degree q − 1. πν Proposition. Let ν be the character of T which has order 2. Then there exist two + + irreducible inequivalent cuspidal representations πν and πν , both of degree (q −1)/2, such that ϕν = χ + + χ − . Furthermore, πν πν (1) χ + (±u1) = χ − (±uε) = ν(±1)(−1 + Γ)/2. πν πν (2) χ − (±u1) = χ + (uε) = ν(±1)(−Γ − 1)/2. πν πν 2 (3) If γ ∈ T and γ 6= 1, then χ + (γ) = χ − (γ) = ν(γ). πν πν × 2 (4) If a ∈ and a 6= 1, then χ + (sa) = χ − (sa) = 0. Fq πν πν B B Proof. We know that r (χ + ) and r (χ − ) are irreducible characters of B having the G πν G πν B property that r (χ + +χ + ) equals χ + +χ + if ν(−1) = 1 and χ − +χ − if ν(−1) = −1. G πν πν σ1 σε σ1 σε This is enought to obtain parts (1) and (2).

49 Part (3) can be proved using orthogonality relations. + − For part (4), since πν and πν are cuspidal, their restrictions to B don’t contain any degree one representation of B. As we have already mentioned, the character of an irreducible representation of B of degree (q − 1)/2 vanishes on elements of the form sa, a2 6= 1. qed

Theorem. Let ψ be a nontrivial character of F . Let λ and ν be the unique characters × P of and T of order 2, respectively. Set Γ = × ψ(x)λ(x). Let τ range over the Fq x∈Fq × characters of Fq whose squares are nontrivial. Let θ range over the characters of T whose squares are nontrivial. Then the irreducible characters of G = SL2(Fq) (for q odd and q 6= 3) are given below. Note that πτ ' πτ −1 and πθ ' πθ−1 . In order, the rows give the values of the characters χ G , the trivial character, the unique irreducible character of iB τ + − + − degree q, χ , χ , χπθ , χ , and χ . πλ πλ πν πν

2 2 ±1 sa, a 6= 1 ±u1 ±uε γ ∈ T, γ 6= 1 τ(±1)(q + 1) τ(a) + τ(a)−1 τ(±1) τ(±1) 0 1 1 1 1 1 q 1 0 0 −1 λ(±1)(q + 1)/2 λ(a) λ(±1)(1 + Γ)/2 λ(±1)(1 − Γ)/2 0 λ(±1)(q + 1)/2 λ(a) λ(±1)(1 − Γ)/2 λ(±1)(1 + Γ)/2 0 θ(±1)(q − 1) 0 −θ(±1) −θ(±1) −θ(γ) − θ(γ)−1 ν(±1)(q − 1)/2 0 ν(±1)(−1 + Γ)/2 ν(±1)(−1 − Γ)/2 ν(γ) ν(±1)(q − 1)/2 0 ν(±1)(−1 − Γ)/2 ν(±1)(−1 + Γ)/2 ν(γ)

In the next chapter we will describe (without proofs) how the above parametrization of the irreducible characters of SL2(Fq) ﬁts in with general results on the characters of ﬁnite groups of Lie type.

50 CHAPTER 4

Representations of ﬁnite groups of Lie type

Let Fq be a finite field of order q and characteristic p. Let G be a finite group of Lie type, that is, G is the Fq-rational points of a connected reductive group G defined over Fq. For example, if n is a positive integer GLn(Fq) and SLn(Fq) are finite groups of Lie type. 0 In Let J = , where In is the n × n identity matrix. Let −In 0

t Sp2n(Fq) = { g ∈ GL2n(Fq) | gJg = J }.

Then Sp2n(Fq) is a symplectic group of rank n and is a ﬁnite group of Lie type. For G = GLn(Fq) or SLn(Fq) (and some other examples), the standard Borel subgroup B of G is the subgroup of G consisting of the upper triangular elements in G.A standard parabolic subgroup of G is a subgroup of G which contains the standard Borel subgroup B. If P is a standard parabolic subgroup of GLn(Fq), then there exists a partition (n1, . . . , nr) of n (a set of positive integers nj such that n1 + ··· + nr = n) such that P = P(n1,...,nr ) =

M n N, where M ' GLn1 (Fq) × · · · × GLnr (Fq) has the form

 A 0 ··· 0    1   0 A2 ··· 0  M =   | A ∈ GL ( ), 1 ≤ j ≤ r .  ......  j nj Fq  . . . .     0 ··· 0 Ar and    In1 ∗ · · · ∗     0 In2 · · · ∗  N =   ,  ......   . . . .     0 ··· 0 Inr where ∗ denotes arbitary entries in Fq. The subgroup M is called a (standard) Levi subgroup of P , and N is called the unipotent radical of P . Note that the partition (1, 1,..., 1) corresponds to B and (n) corresponds to G. A standard parabolic subgroup of SLn(Fq) is equal to P(n1,...,nr ) ∩ SLn(Fq) for some partition (n1, . . . , nr) of n.A parabolic subgroup of G is a subgroup which is conjugate to a standard parabolic subgroup. By replacing Fq by a field F in the above definitions, we can define parabolic subgroups of GLn(F ). For more information on parabolic subgroups of general linear groups see the book of Alperin and Bell.

For convenience, assume that G = GLn(Fq). Let P = P(n1,...,nr ) be a proper standard parabolic subgroup of G. Let πj be an irreducible representation of GLnj (Fq), 1 ≤ j ≤ r.

51 Then π := π1 ⊗· · ·⊗πr is an irreducible representation of M. Extend π to a representation G G of P by letting elements of N act via the identity (on the space of π). Let iP π = IndP π be the representation of G induced from π. This process of going from a representation of a Levi subgroup of a proper parabolic subgroup to a representation of G is known as parabolic induction (or Harish-Chandra induction). It is possible to show that if P 0 is another parabolic subgroup of G having M as a Levi subgroup, and unipotent radical N 0, G G G G then iP π ' iP 0 π. For this reason, the notation iM π is sometimes used in place of iP π. Now we can deﬁne a standard parabolic subgroup PM of M to be a group of the

form PM = P1 × · · · × Pr where Pj is a standard parabolic subgroup of GLnj (Fq). Write Pj = Mj n Nj, where Mj is the standard Levi subgroup of Pj and Nj is the unipotent radical of Pj. Let NM = N1 × · · · Nr. Then PM N is a standard parabolic subgroup of

GLnj (Fq), with standard Levi factor M1 × · · · × Mr and unipotent radical NM N. Let σ = σ1 ⊗ · · · ⊗ σr where σj is an irreducible representation of Mj. Then we can check that iG σ ' iG(iM σ), where on the left side, σ is extended to P N by letting elements of PM N P PM M NM N act trivially, and on the right side, σ is extended to PM by letting NM act trivially. This property is called transitivity of parabolic induction. Let N be the unipotent radical of a proper parabolic subgroup of G and let π be a N representation of G. Then the restriction rG π = πN of π to N is a direct sum of irreducible representations of N. Via Frobenius reciprocity type arguments, we can prove that if an irreducible representation π of G contains the trivial representation of the unipotent radical of proper parabolic subgroup of G, then π occurs as a subrepresentation of a parabolically N induced representation of G. An irreducible representation π of G is cuspidal if rG π does not contain the trivial representation of N for all choices of unipotent radicals of proper G 0 parabolic subgroups of G. If π is cuspidal, then by Frobenius reciprocity, HomG(π, iP π ) = 0 if P is a proper parabolic subgroup of G and π0 is an irreducible representation of a Levi factor of P . The following theorem is valid for irreducible representations of ﬁnite groups of Lie type, not just for general linear groups.

Theorem. (Proposition 9.13 of Carter’s book) Let π be an irreducible representation of G. Then one of the following holds: (1) π is cuspidal (2) There exists a proper parabolic subgroup P = M n N of G and a cuspidal represen- 0 G 0 tation π of M such that HomG(π, iP π ) 6= 0.

As a consequence of the above theorem, one approach to ﬁnding the irreducible representations of G involves two steps: Step 1 : Find all cuspidal representations of those groups occurring as Levi factors of parabolic subgroups of G (including G itself). G 0 Step 2 : Find all of the irreducible constituents of the representations iP π where P is a

52 proper parabolic subgroup of G and π0 is a cuspidal representation of a Levi factor of P . In SL2(Fq), there is one conjugacy class of proper parabolic subgroups, namely the × conjugacy class of the standard Borel subgroup. And the subgroup A ' Fq of B is a standard Levi factor. Every irreducible representation (one-dimensional representation) of A is cuspidal, since A has no proper parabolic subgroups. The above theorem tells us that the non-cuspidal representations of GL2(Fq) are exactly the irreducible subrepresentations G of the representations iBτ as τ ranges over the set of one-dimensional representations of A. × As we saw in Chapter 3, letting ε be a ﬁxed nonsquare in Fq , the cuspidal representations of SLn(Fq) are associated to the nontrivial characters of the group a bε T = | a, b ∈ , a2 − b2ε = 1 . b a Fq

In the case of SL2(Fq) the groups A and T are representatives for the conjugacy classes of maximal tori in SL2(Fq). In 1976, for G a ﬁnite group of Lie type, Deligne and Lusztig published a paper showing that it is possible to associate a virtual character of G to each character of a maximal torus (see references). Next we describe the maximal tori in GLn(Fq) and SLn(Fq). Let k be a positive integer. The Fqk is a vector space over Fq of dimension k. Choose a basis β = { x1, . . . , xk } of Fqk over Fq. Given y ∈ Fqk , let gy ∈ GLk(Fq) be the matrix relative to β of the linear × transformation on Fqk given by left multiplication by y. Then { gy | y ∈ Fqk } is a subgroup × of GLk(Fq) which is isomorphic, via y 7→ gy, to Fqk . This subgroup depends on the choice of basis β. Any two such subgroups of GLk(Fq) are conjugate, via the relevant change of basis matrix.

Suppose that n1, . . . , nr are integers such that n1 ≥ n2 ≥ · · · ≥ nr > 0 and n1 + ··· + × × n nr = n. Fix a basis of Fq j over Fq, 1 ≤ j ≤ r. Then the group Fqn1 × · · · × Fqnr , is isomorphic to a subgroup T(n1,...,nr ) of GLn1 (Fq)×· · ·×GLnr (Fq) ⊂ GLn(Fq). A maximal torus T in GLn(Fq) is a subgroup which is conjugate to T(n1,...,nr ) for some (n1, . . . , nr) as above. A maximal torus in SLn(Fq) is conjugate to some T(n1,...,nr ) ∩ SLn(Fq). For SL2(Fq) there are 2 conjugacy classes of maximal tori, represented by T(1,1) ∩SL2(Fq) = A, and T(2) ∩ SL2(Fq), the group T mentioned above. Let T be a maximal torus in G and let θ be a character of T . Let RT,θ be the virtual character of G associated to the pair (T, θ) by Deligne and Lusztig. The value of RT,θ on an element g of G is given by an alternating sum of the trace of the action of g on certain `-adic cohomology groups. For more information on this, see the paper of Deligne and Lusztig, or the book of Carter. A matrix u ∈ GLn(Fq) is unipotent if u − 1 is nilpotent. A matrix γ ∈ GLn(Fq) is semisimple if γ is diagonalizable over some ﬁnite extension of Fq. Let g ∈ GLn(Fq). The

53 characteristic polynomial of g, being a polynomial of degree n, splits over a ﬁnite extension

Fqk of Fq, for some k ≤ n. Results from linear algebra tell us that there exist matrices γ and u ∈ GLn(Fqk ) such that γ is diagonalizable and u is unipotent, with γu = uγ = g. It can be shown that γ, u ∈ GLn(Fq). Hence any element in GLn(Fq) can be expressed in a unique way in the form g = γu, with γ ∈ GLn(Fq) semisimple, u ∈ GLn(Fq) unipotent, and γu = uγ. This is called the (multiplicative) Jordan decomposition of g. If G is a ﬁnite group of Lie type, then G is a subgroup of GLn(Fq) for some n, and if g = γu is the Jordan decomposition of g in GLn(Fq), the elements γ and u lie in G. So there is a Jordan decomposition for elements of a ﬁnite group of Lie type. G Attached to any a maximal torus T in G, there exists a particular class function QT G on G, called the Green function corresponding to T . The Green function QT is supported on the unipotent set. If G = GLn(Fq), the values of the Green functions are known. In G fact, if u is unipotent the value QT (u) is obtained as the value of a certain polynomial (depending on T and clG(u)) in the variable q.

Theorem (Character formula for RT,θ). Let T be a maximal torus in G = GLn(Fq) or SLn(Fq) and let θ be a character of T . Let g ∈ GLn(Fq) have Jordan decomposition g = γu, with semisimple part γ and unipotent part u. Let H be the centralizer of γ in G. Then −1 X −1 H RT,θ(g) = |H| θ(x γx)QxT x−1 (u). x∈G, x−1γx∈T

Exercises: Let γ ∈ GLn(Fq) be semisimple.

(1) Prove that some conjugate of γ lies in T(n1,...,nr ) for some (not necessarily unique) (n1, . . . , nr). (2) Prove that the centralizer H of γ in G is isomorphic to a direct product of the form

s s GLr1 (Fq 1 ) × · · · × GLrt (Fq t ), where r1s1 + ··· + rtst = n.

The character formula for RT,θ resembles the Frobenius character formula in certain ways. If the conjugacy class of the semisimple part γ of g does not intersect T , then

RT,θ(g) = 0. And when the conjugacy class of γ does intersect T , the expression for

RT,θ(g) involves values of θ on certain conjugates of γ in T . In the special case where

T = T(1,...,1) or T(1,...,1) ∩ SLn( q), then it can be shown that RT,θ = ±χ G . F iB θ If T and T 0 are maximal tori in G, let

N(T,T 0) = { g ∈ G | gT g−1 = T 0 } W (T,T 0) = { T g | g ∈ N(T,T 0) }

0 Theorem (Orthogonality relations for RT,θ’s). Let θ and θ be characters of T and T 0, respectively. Then

0 ω 0 hRT,θ,RT 0,θi = { ω ∈ W (T,T ) | θ = θ },

54 where ωθ0(γ) := θ0(gγg−1), for γ ∈ T and ω = T g.

0 Corollary. If T and T are not G-conjugate, then hRT,θ,RT 0,θ0 i = 0.

Note that this corollary does not tell us that the set of irreducible characters appear- 0 ing in RT,θ is disjoint from those appearing in RT 0,θ0 when T and T are not conjugate. Consider the example of G = SL2(Fq). Let τ be a character of the maximal torus A in G G = SL2(Fq). According to comments above, RA,τ = iBτ. Let T be as in Chapter 3. Then it is possible to show that if θ is a character of T and ϕθ is as in Chapter 3, then RT,θ = ±ϕθ. Now suppose that τ = τ0 is the trivial character of A and θ = θ0 is the trivial character of T . Let χq be the character of the irreducible representation of SL2(Fq) of degree q. Let χ0 be the character of the trivial representation. As we saw in Chapter 3,

χ G = RA,τ = χ0 + χq and ϕθ = −χ0 + χq. iB τ0 0 −1 Let NG(T ) = { g ∈ G | gT g = T } be the normalizer of T in G. Then W (T ) := w NG(T )/T is a ﬁnite group. A character θ of T is said to be in general position if θ 6= θ for all nontrivial elements of W (T ).

Theorem (corollary of above theorem). If θ is in general position, then ±RT,θ is an irreducible character of G.

Theorem. If π is an irreducible representation of G, then there exists a maximal torus T of G and a character θ of T such that hχπ,RT,θi= 6 0.

Suppose that (n1, . . . , nr) 6= (n). Then T(n1,...,nr ) is a subgroup of the Levi factor

M = GLn1 (Fq)×· · ·×GLnr (Fq) of a proper parabolic subgroup P of G. Any class function f on M can be extended to a class function on P = M n N by setting f(mu) = f(m), G m ∈ M, u ∈ N. Hence we can view iP as a map from class functions on M to class functions on G. Deligne and Lusztig proved that if T = T(n1,...,nr ) and θ is a character of G M M T , then RT,θ = iP (RT,θ), where RT,θ is the virtual character of M corresponding to the pair (T, θ). (Note that T is a maximal torus in M). It follows that RT,θ is in the span of the G irreducible characters of G which occur as constituents of representations iP σ, for various representations σ of M. Thus hχπ,RT,θi = 0 for all irreducible cuspidal representations π.

Proposition. Suppose that π is an (irreducible) cuspidal representation of GLn(Fq), n ≥ 2. Then

(1) hχπ,RT(n),θi= 6 0 for some character θ of T(n).

(2) χπ = ±RT(n),θ for some character θ of T(n) that is in general position. (3) If T is a maximal torus which is not conjugate to T(n) and θ is a character of T , then hχπ,RT,θi = 0.

Parts (1) and (3) have analogues for cuspidal representations for other ﬁnite groups of

Lie type. In those cases, the maximal torus T(n) must be replaced by a set of representatives

55 for the conjugacy classes of maximal tori in G that do no intersect the Levi factor of any proper parabolic subgroup of G. The analogue of Part (2) does not hold for all cuspidal representations of other ﬁnite groups of Lie type. The two irreducible characters of of SL2(Fq) of degree (q − 1)/2 which were produced in Chapter 3 correspond to cuspidal representations whose characters are not of the form ±RT,θ (for any choice of T and θ). All of the irreducible characters of SL2(Fq) of degree q − 1 correspond to cuspidal 2 representations whose characters equal ±RT,θ (with θ 6= 1, and T as in Chapter 3). Hecke algebras are often used in the study of the representations of ﬁnite groups of Lie type. Suppose that P and P 0 are parabolic subgroups of G and (π, V ) and (π0,V 0) are irreducible representations of the Levi factors M and M 0 of standard parabolic subgroups P and P 0, respectively. Then, according to results from Chapter 2, the space G G 0 0 HomG(iP π, iP 0 π ) is isomorphic to the space of functions ϕ : G → EndC(V,V ) such that ϕ(xgx0) = π(x)◦ϕ(g)◦π0(x0) for all g ∈ G, x ∈ P , and x0 ∈ P 0. In view of results discussed above, the case where π and π0 are cuspidal is of particular interest. The following theorem is proved by studying properties of the above isomorphism - in particular, by analyzing the properties of the functions supported on each double coset P gP 0 and satisfying the above conditions.

Theorem. Let π and π0 be cuspidal representations of Levi factors M and M 0 of standard 0 parabolic subgroups P and P of GLn(Fq), respectively. Then 0 G G 0 (1) If M and M are not conjugate, then HomG(iP π, iP 0 π ) = 0. 0 G G 0 G G 0 (2) If M and M are conjugate, then either HomG(iP π, iP 0 π ) = 0 or iP π ' iP 0 π . Suppose that P 0 = P and π0 = π. Then H(G, π) is a Hecke algebra which is iso- G G morphic (as an algebra) to HomG(iP π, iP π). It is possible to prove that decomposition G of representations of the form iP π into direct sums of irreducible representations can be reduced to cases where P = P(m,...,m) and m 6= n is a divisor of n (with m occurring n/m times). In those cases, π = σ ⊗ · · · ⊗ σ for some cuspidal representation σ of GLm(Fq) 0 0 (where σ occurs n/m times in the tensor product). Let G = GLn/m(Fqm ). Let B be the standard Borel subgroup of G0. Then, letting π0 be the trivial representation of G0, we 0 0 G0 0 G0 0 know that the Hecke algebra H(G , π ) is isomorphic to HomG0 (iB0 π , iB0 π ). Theorem. (Notation as above). The algebra H(G, π) is isomorphic (in a canonical way) to the Hecke algebra H(G0, π0).

Corollary. There is a canonical bijection τ ↔ τ 0 between the set of irreducible con- G 0 G0 0 stituents τ of iP π and the set of irreducible constituents of τ of iB0 π . The bijection satisifes: P 0 (1) The multiplicity of π in rGτ equals the multiplicity of the trivial representation π of 0 B0 B in rG0 τ.

56 (2) The degree of τ divided by the degree of τ 0 equals the degree of π times |G||P |−1|G0|−1|B0|.

57 CHAPTER 5

Topological Groups, Representations, and Haar Measure

5.1. Topological spaces If X is a set, a family U of subsets of X defines a topology on X if (i) ∅ ∈ U, X ∈ U. (ii) The union of any family of sets in U belongs to U. (iii) THe intersection of a finite number of sets in U belongs to U. If U defines a topology on X, we say that X is a topological space. The sets in U are called open sets. The sets of the form X \ U, U ∈ U, are called closed sets. If Y is a subset of X the closure of Y is the smallest closet set in X that contains Y .

Let Y be a subset of a topological space X. Then we may deﬁne a topology UY on Y , called the subspace or relative topology, or the topology on Y induced by the topology on X, by taking UY = { Y ∩ U | U ∈ U }. A system B of subsets of X is called a basis (or base) for the topology U if every open set is the union of certain sets in B. Equivalently, for each open set U, given any point x ∈ U, there exists B ∈ B such that x ∈ B ⊂ U. Example: The set of all bounded open intervals in the real line R forms a basis for the usual topology on R.

Let x ∈ X.A neighbourhood of x is an open set containing x. Let Ux be the set of all neighbourhoods of x. A subfamily Bx of Ux is a basis or base at x, a neighbourhood basis at

x, or a fundamental system of neighbourhoods of x, if for each U ∈ Ux, there exists B ∈ Bx such that B ⊂ U. A topology on X may be speciﬁed by giving a neighbourhood basis at every x ∈ X. If X and Y are topological spaces, there is a natural topology on the Cartesian product X × Y that is deﬁned in terms of the topologies on X and Y , called the product topology.

Let x ∈ X and y ∈ Y . The sets Ux × Vy, as Ux ranges over all neighbourhoods of x, and Vy ranges over all neighbourhoods of y forms a neighbourhood basis at the point (x, y) ∈ X × Y (for the product topology). If X and Y are topological spaces, a function f : X → Y is continuous if whenever U is an open set in Y , the set f −1(U) = { x ∈ X | f(x) ∈ U } is an open set in X.A function f : X → Y is a homeomorphism (of X onto Y ) if f is bijective and both f and f −1 are continuous functions. An open covering of a topological space X is a family of open sets having the property that every x ∈ X is contained in at least one set in the family. A subcover of an open covering is a an open covering of X which consists of sets belonging to the open covering. A topological space X is compact if every open covering of X contains a ﬁnite subcover.

58 A subset Y of a topological space X is compact if it is compact if Y is compact in the subspace topology. A topological space X is locally compact if for each x ∈ X there exists a neighbourhood of x whose closure is compact.

A topological space X is Hausdorﬀ (or T2) if given distinct points x and y ∈ X, there exist neighbourhoods U of x and V of y such that U ∩ V = ∅. A closed subset of a locally compact Hausdorﬀ space is locally compact.

5.2. Topological groups A topological group G is a group that is also a topological space, having the property −1 the maps (g1, g2) 7→ g1g2 from G×G → G and g 7→ g from G to G are continuous maps. In this definition, G × G has the product topology. Lemma. Let G be a topological group. Then (1) The map g 7→ g−1 is a homeomorphism of G onto itself. −1 (2) Fix g0 ∈ G. The maps g 7→ g0g, g 7→ gg0, and g 7→ g0gg0 are homeomorphisms of G onto itself. A subgroup H of a topological group G is a topological group in the subspace topology. Let H be a subgroup of a topological group G, and let p : G → G/H be the canonical mapping of G onto G/H. We define a topology UG/H on G/H, called the quotient topology, by UG/H = { p(U) | U ∈ UG }. (Here, UG is the topology on G). The canonical map p is open (by definition) and continuous. If H is a closed subgroup of G, then the topological space G/H is Hausdorff. If H is a normal subgroup of G, then G/H is a topological group. If G and G0 are topological groups, a map f : G → G0 is a continuous homomorphism of G into G0 if f is a homomorphism of groups and f is a continuous function. If H is a closed normal subgroup of a topological group G, then the canonical mapping of G onto G/H is an open continuous homomorphism of G onto G/H. A topological group G is a locally compact group if G is locally compact as a topological space. Proposition. Let G be a locally compact group and let H be a closed subgroup of G. Then (1) H is a locally compact group (in the subspace topology). (2) If H is normal in G, then G/H is a locally compact group. (3) If G0 is a locally compact group, then G×G0 is a locally compact group (in the product topology).

5.3. General linear groups and matrix groups Let F be a ﬁeld that is a topological group (relative to addition). Assume that points in F are closed sets in the topology on F . For example, we could take F = R,

59 C or the p-adic numbers Qp, p prime. Let n be a positive integer. The space Mn×n(F ) of n × n matrices with entries in F is a topological group relative to addition, when n2 Mn×n(F ) ' F is given the product topology. The multiplicative group GLn(F ), being a subset (though not a subgroup) of Mn×n(F ), is a topological space in the subspace topology. The determinant map det from Mn×n(F ) to F , being a polynomial in matrix entries, is a continuous function. Now F × = F \{0} is an open subset of F (since points are −1 × closed in F ). Therefore, by continuity of det, GLn(F ) = det (F ) is an open subset of

Mn×n(F ). It is easy to show that matrix multiplication, as a map (not a homomorphism) from Mn×n(F ) × Mn×n(F ) to Mn×n(F ) is continuous. It follows that the restriction to GLn(F ) × GLn(F ) is also continuous. Let g ∈ GLn(F ). Recall that Cramer’s rule gives a formula for the ijth entry of g−1 as the determinant of the matrix given by deleting the ith row and jth column of g, divided by det g. Using this, we can prove that g 7→ g−1 is a continuous map from GLn(F ) to GLn(F ). Therefore GLn(F ) is a topological group. We can also see that if F is a locally compact group (for example if F = R, C or Qp), then GLn(F ) is a locally compact group.

The group SLn(F ), being the kernel of the continuous homomorphism det : GLn(F ) → × F , is a closed subgroup of SLn(F ), so is a locally compact group whenever F is a lo- 0 In cally compact group. If In is the n × n identity matrix and J = , then −In 0 t Sp2n(F ) = { g ∈ GL2n(F ) | gJg = J } is a closed subgroup of GL2n(F ). If S ∈ GLn(F ) t t is a symmetric matrix, that is S = S, the group On(S) = { g ∈ GLn(F ) | gSg = S } is an orthogonal group, and is a closed subgroup of GLn(F ). Depending on the ﬁeld F , diﬀerent choices of S can give rise to non-isomorphic orthogonal groups. If E is a quadratic extension of F and X ∈ Mn×n(E), let X¯ be the matrix obtained from X by letting the nontrivial element of the Galois group Gal(E/F ) act on each the entries of X. t¯ Suppose that h ∈ GLn(E) is a matrix such that h = h (h is hermitian). Then the group t U(h) = { g ∈ GL2n(E) | ghg¯ = h } is called a unitary group and is a closed subgroup of GLn(E). If (n1, . . . , nr) is a partition of n then the corresponding standard parabolic subgroup P = P(n1,...,nr ) of GLn(F ) is a closed subgroup of GLn(F ), as are any Levi factor of P , and the unipotent radical of P .

5.4. Matrix Lie groups A Lie group is a topological group that is a differentiable manifold with a group structure in which the multiplication and inversion maps from G×G to G and from G to G are smooth maps. Without referring to the differentiable manifolds, we may define a matrix Lie group, or a closed Lie subgroup of GLn(C) to be a closed subgroup of the topological group GLn(C). (This latter definition is reasonable because GLn(C) is a Lie group, and it can be shown that a closed subgroup of a Lie group is also a Lie group). A connected matrix Lie group is reductive if it is stable under conjugate transpose, and semisimple

60 if it is reductive and has ﬁnite centre. The book of Hall [Hall] gives an introduction to matrix Lie groups, their structure, and their ﬁnite-dimensional representations. For other references on Lie groups and their representations, see [B], [K1] and [K2].

5.5. Finite-dimensional representations of topological groups and matrix Lie groups

Let G be a topological group. A (complex) finite-dimensional continuous representation of G is a finite-dimensional (complex) representation (π, V ) of G having the property that the map g 7→ [π(g)]β from G to GLn(C) is a continuous homomorphism for some (hence any) basis β of V . The continuity property is equivalent to saying that every matrix coefficient of π is a continuous function from G to C. Hence to prove the following lemma, we need only observe that the character of π is a finite sum of matrix coefficients of π.

Lemma. Let π be a continuous ﬁnite-dimensional representation of G. Then the character

χπ of π is a continuous function on G. Example: Let π be a continuous one-dimensional representation of the locally compact group R. Then π is a continuous function from R to C such that π(0) = 1 and π(t1 + t2) = π(t1)π(t2) for all t1, t2 ∈ R. If f : R → C is continuously diﬀerentiable and the support R ∞ of f is contained in a compact subset of R, then −∞ f(t)π(t) dt converges. Choose f so R ∞ that c = −∞ f(t)π(t) dt 6= 0. Multiplying π(t1 + t2) by f(t2) and integrating, we have

Z ∞ Z ∞ f(t2)π(t1 + t2) dt2 = π(t1) f(t2)π(t2) dt2 = cπ(t1), t1 ∈ R. −∞ −∞

Then Z ∞ Z ∞ −1 −1 π(t1) = c f(t2)π(t1 + t2) dt2 = c π(t)f(t − t1) dt, t1 ∈ R. −∞ −∞

R ∞ Because t1 7→ −∞ π(t)f(t − t1) dt is a differentiable function of t1, we see that π is a differentiable function. Differentiating both sides of π(t1 + t2) = π(t1)π(t2) with respect 0 0 0 to t1 and then setting t1 = 0 and t = t2, we obtain π (t) = π (0)π(t). Setting k = π (0), we have π0(t) = kπ(t), t ∈ R. Solving this differential equation yields π(t) = aekt for some a ∈ C. And π(0) = 1 forces a = 1. Hence π(t) = ekt. Now if we take a z ∈ C, it is clear that t 7→ ezt is a one-dimensional continuous representation of R.

zt Lemma. Let z ∈ C. Then πz(t) = e deﬁnes a one-dimensional continuous representation of R. The representation πz is unitary if and only if the real part of z equals 0. Each one-dimensional continuous representation of R is of the form πz for some z ∈ C, and

61 any one-dimensional continuous representation of R is a smooth (inﬁnitely diﬀerentiable) function of t. Theorem. A continuous homomorphism from a Lie group G to a Lie group G0 is smooth.

Let G be a Lie group (for example, a matrix Lie group). Then, because GLn(C) is a Lie group, via a choice of basis for the space of the representation, a finite-dimensional representation of G is a continuous homomorphism from G to GLn(C). According to the theorem, the representation must be a smooth map from G to GLn(C). Corollary. A continuous finite-dimensional representation of a Lie group is smooth. Definition. A Lie algebra over a field F is a vector space g over F endowed with a bilinear map, the Lie bracket, denoted (X,Y ) 7→ [X,Y ] ∈ g satisfying [X,Y ] = −[Y,X] and the Jacobi identity

[X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0 ∀ X,Y,Z ∈ g.

If G is a Lie group, the Lie algebra g of G is defined to be the set of left-G-invariant smooth vector fields on G. A vector field is a smoothly varying family of tangent vectors, one for each g ∈ G, and it can be shown that if X is identified with the corresponding tangent vector at the identity element, then the Lie algebra g is identified with the tangent space at the identity. If we work with matrix Lie groups, we can take a different approach. If X ∈ Mn×n(C), X P∞ k then the matrix exponential e = k=0 X /k! is an element of GLn(C). Proposition. Let G be a matrix Lie group. Then the Lie algebra g is equal to

tX g = { X ∈ Mn×n(C) | e ∈ G ∀ t ∈ R }.

and the bracket [X,Y ] of two elements of g is equal to the element XY −YX of Mn×n(C). X tr X Using the fact that det(e ) = e , we can see that the Lie algebra sln(C), resp. sln(R), of SLn(C), resp. SLn(R), is just the set of matrices in Mn×n(C), resp. Mn×n(R), that have trace equal to 0. Now suppose that G = Sp2n(F ) with F = R or C. Note that t etX ∈ G if and only if JetX J −1 = e−tX . From

d tX tX X = (e ) |t=0 = lim(e − 1)/t. dt t→0

t we can see that JetX J −1 = e−tX for all t ∈ R implies JXtJ −1 = −X. The converse is easy to see. Therefore the Lie algebra sp2n(F ) of Sp2n(F ) is given by

t −1 t sp2n(F ) = { X ∈ Mn×n(F ) | JX J = −X } = { X ∈ Mn×n(F ) | JX + XJ = 0 }.

62 The same type of approach can be used to ﬁnd the Lie algebras of orthogonal and unitary matrix Lie groups.

If V is a ﬁnite-dimensional complex vector space, EndC(V ) is a Lie algebra relative to the bracket [X,Y ] = X ◦ Y − Y ◦ X. This Lie algebra is denoted by gl(V ). A linear map φ : g → gl(V ) is a Lie algebra homomorphism if

φ([X,Y ]) = φ(X) ◦ φ(Y ) − φ(Y ) ◦ φ(X),X,Y ∈ g.

A finite-dimensional representation of g is a Lie algebra homomorphism from g to gl(V ) for some finite-dimensional complex vector space V . Proposition. Let G be a matrix Lie group, and let (π, V ) be a continuous finite-dimensional representation of G. Then there is a unique representation dπ of of the Lie algebra g of G (acting on the space V ) such that

π(eX ) = edπ(X),X ∈ g.

d tX Furthermore dπ(X) = dt π(e ) |t=0, X ∈ g, and π is irreducible if and only if dπ is irreducible. If the matrix Lie group G is simply connected, that is, the topological space G is simply connected, then any ﬁnite-dimensional representation of g lifts to a ﬁnite-dimensional representation of G, and the representations of G and g are related as in the above proposition.

5.6. Groups of t.d. type A Hausdorff topological group is a t.d. group if G has a countable neighbourhood basis at the identity consisting of compact open subgroups, and G/K is a countable set for every open subgroup K of G. Some t.d. groups are matrix groups over p-adic fields. Let p be a prime. Let x ∈ Q×. Then there exist unique integers m, n and r such that m and n are nonzero and relatively prime, p does not divide m or n, and x = prm/n. Set −r × |x|p = p . This defines a function on Q , which we extend to a function from Q to the set of nonnegative real numbers by setting |0|p = 0. The function | · |p is called the p-adic absolute value on Q. It is a valuation on Q - that is, it has the properties (i) |x|p = 0 if and only if x = 0 (ii) |xy|p = |x|p|y|p

(iii) |x + y|p ≤ |x|p + |y|p. The usual absolute value on the real numbers is another example of a valuation on Q. The p-adic abolute value satisﬁes the ultrametric inequality, that is, |x + y|p ≤ max{|x|p, |y|p}. Note that the ultrametric inequality implies property (iii) above. A valuation that satisﬁes the ultrametric inequality is called a nonarchimedean valuation.

63 × × Note that the set { |x|p | x ∈ Q } is a discrete subgroup of R . Hence we say that |·|p is a discrete valuation. The usual absolute value on Q is an example of an archimedean valuation. Clearly it is not a discrete valuation. Two valuations on a ﬁeld F are said to be equivalent is one is a positive power of the other.

Theorem. (Ostrowski) A nontrivial valuation on Q is equivalent to the usual absolute value or to | · |p for some prime p. If F is a field and | · | is a valuation on F , the topology on F induced by | · | has as a basis the sets of the form U(x, ) = { y ∈ F | |x − y| < }, as x varies over F , and varies over all positive real numbers. A field F 0 with valuation | · |0 is a completion of the field F with valuation | · | if F ⊂ F 0, |x|0 = |x| for all x ∈ F , F 0 is complete with respect to | · |0 (every Cauchy sequence with respect to | · |0 has a limit in F 0) and F 0 is the closure of F with respect to | · |. So F 0 is the smallest field containing F such that F 0 is complete with respect to | · |0. The real numbers is the completion of Q with respect to the usual absolute value on Q. The p-adic numbers Qp is the completion of Q with respect to | · |p. (We denote the extension of | · |p to Qp by | · |p also). The p-adic integers Zp is the set { x ∈ Qp | |x|p ≤ 1 }. Note that Zp is a subring of Qp (this follows from the ultrametric inequality and the mulitiplicative property of |·|p), and Zp contains Z. The set pZp (the ideal of Qp generated by the element p) is a maximal ideal of Zp and Zp/pZp is therefore a field. Let x ∈ Q×. Write x = prm/n with r ∈ Z and m and n nonzero integers such that m and n are relatively prime and not divisible by p. Because m and n are relatively prime and not divisible by p, the equation nX ≡ m(mod p) has a unique solution ar ∈ { 1, . . . , p −1 }.

That is, there is a unique integer ar ∈ {1, . . . , p − 1 } such that p divides m − nar. Since r |n|p = 1, p divides m − nar is equivalent to |(m/n) − ar|p < 1, and also to |x − arp |p < −r r s 0 0 0 0 |x|p = p . Expressing x − arp in the form p m /n with s > r and m and n relatively prime integers, we repeat the above argument to produce an integer as ∈ {1, . . . , p − 1} such that r s r −s |x − arp − asp |p < |x − arp |p = p .

If s > r + 1, set ar+1 = ar+2 = ··· as−1 = 0, to get s X n −s |x − anp |p < p . n=r

Continuing in this manner, we see that there exists a sequence { an | n ≥ r } such that

an ∈ { 0, 1, . . . , p − 1 } and, given any integer M ≥ r,

M X n −M |x − anp |p < p . n=r

64 P∞ n It follows that n=r anp converges in the p-adic topology to the rational number x. On the other hand, it is quite easy to show that if an ∈ { 0, 1, . . . , p − 1} and r is P∞ n an integer, then n=r anp converges to an element of Qp (though not necessarily to a rational number).

P∞ n Lemma. A nonzero element x of Qp is uniquely of the form n=r anp , with an ∈ −r { 0, 1, 2, . . . , p − 1}, for some integer r with ar 6= 0. Furthermore, |x|p = p . (Hence x ∈ Zp if and only if r ≥ 0).

Lemma. Zp/pZp ' Z/pZ.

P∞ n Proof. Let a ∈ Zp. According to the above lemma, a = n=r anp for some sequence −r { an | n ≥ r }, where |a|p = p ≤ 1 implies that r ≥ 0. If r > 0, then a ∈ pZp. For convenience, set a0 = 0 when |a|p < 1. If r = 0, then a0 ∈ { 1, . . . , p − 1 }. Define a map from Zp to Z/pZ by a 7→ a0. This is a surjective ring homomorphism whose kernel is equal to pZp. qed A local field F is a (nondiscrete) field F which is locally compact and complete with respect to a nontrivial valuation. The fields R, C, and Qp, p prime, are local fields. If | · | is a nontrivial nonarchimedean valuation on a field F , then { x ∈ F | |x| < 1} is a maximal ideal in the ring { x ∈ F | |x| ≤ 1 }, so the quotient is a field, called the residue class field of F . The following lemma can be used to check that Qp is a local field.

Lemma. Let | · | be a nonarchimedean valuation on a field F . Then F is locally compact with respect to | · | if and only if (1) F is complete (with respect to | · |) (2) | · | is discrete (3) The residue class field of F is finite.

N For every integer N, p Zp is a compact open (and closed) subgroup of Qp. It is not N hard to see that { p Zp | N ≥ 0 } forms a countable neighbourhood basis at the identity × × element 0. From the above lemma, we have that Qp ' hpi × Zp . Hence Qp/Zp is discrete. N It can be shown that any open subgroup of Qp is of the form p Zp for some integer N. Thus Qp/K is discrete for every open subgroup K of Qp. So the group Qp is a t.d. group. For more information on valuations, the p-adic numbers, and p-adic ﬁelds, see the beginning of [M] (course notes for Mat 1197).

As discussed in the section 5.3, because Qp is locally compact, the topological group GLn(Qp) is also locally compact. In fact GLn(Qp) is a t.d. group. If j is a positive integer, j let Kj be the set of g ∈ GLn(Qp) such that every entry of g − 1 belongs to p Zp. Then Kj is a compact open subgroup, and { Kj | j ≥ 1 } forms a countable neighbourhood basis at the identity element 1.

65 Let K be an open subgroup of GLn(Qp). Then Kj ⊂ K for some j ≥ 1. Hence to prove that G/K is countable, it suﬃces to prove that G/Kj is countable for every j. For

a discussion of the proof that G/Kj is countable, see [M]. Closed subgroups (and open subgroups) of t.d. groups are t.d. So any closed subgroup of GLn(Qp) is a t.d. group. These groups are often called p-adic groups. We remark that groups like GLn(Qp), SLn(Qp), Sp2n(Qp), etc., are the groups of Qp-rational points of reductive linear algebraic groups that are defined over Qp. Such groups have another topology, the Zariski topology (coming from the variety that is the algebraic group). The structure of these groups is often studied via algebraic geometry, in contrast with the structure of Lie groups, which is studied via differential geometry. As with Lie groups, there is a notion of smoothness for representations. A (complex) representation (π, V ) is smooth if for each v ∈ G, the subgroup { g ∈ G | π(g)v = v } is an open subgroup of G. This definition is also valid if V is infinite-dimensional. This notion of smoothness is very different from that for Lie groups - in fact, connected Lie groups don’t have any proper open subgroups. Because of the abundance of compact open subgroups in t.d. groups, and the fact that the general theory of representations of compact groups is well understood (see Chapter 6), properties of representations of t.d. groups are often studied via their restrictions to compact open subgroups.

Lemma. Suppose that (π, V ) is a smooth ﬁnite-dimensional representation of a compact t.d. group G. Then there exists an open compact normal subgroup K of G and a representation ρ of the ﬁnite group G/K such that ρ(gK)v = π(g)v for all g ∈ G and v ∈ K.

Proof. By smoothness of π and finite-dimensionality of π, there exists an open compact 0 0 0 0 subgroup K of G such that π(k )v = v for all k ∈ K and v ∈ V . Choose a set { g1, . . . , gr } 0 r 0 −1 of coset representatives for G/K . The subgroup K := ∩j=1kjK kj is an open compact normal subgroup of G and π(k)v = v for all k ∈ K and v ∈ V . It follows that there exists a representation (ρ, V ) of the finite group G/K such that ρ(gK)v = π(g)v, v ∈ V , and g ∈ G. qed We remark that the subgroup K0 (hence the representation ρ) in the above lemma are not unique. Now suppose that (π, V ) is a smooth (not necessarily finite-dimensional) representation of a (not necessarily compact) t.d. group G. Let K be a compact open K subgroup of G. The restriction πK = rG π of π to K is a (possibly infinite) direct sum of irreducible smooth representations of K. As we will see in Chapter 6, irreducible unitary representations of compact groups are finite-dimensional. Applying the above lemma, we

can see that each of the irreducible representations of K which occurs in πK is attached to a representation of some finite group. One difficulty in studying the representations of non-compact t.d. groups involves determining which compact open subgroups K and which irreducible constituents of πK can be used to effectively study properties of π. For

66 more information on representations of p-adic groups, see the notes [C] or the course notes for Mat 1197 ([M])

5.7. Haar measure on locally compact groups If X is a topological space, a σ-ring in X is a nonempty family of subsets of X having the property that arbitrary unions of elements in the family belong to the family, and if A and B belong to the family, then so does { x ∈ A | x∈ / B }. If X is a locally compact topological space, the Borel ring in X is the smallest σ-ring in X that contains the open sets. The elements of the Borel ring are called Borel sets. A function f : X → R is (Borel) measurable if for every t > 0, the set { x ∈ X | |f(x)| < t } is a Borel set. Let G be a locally compact topological group. A left Haar measure on G is a nonzero regular measure µ` on the Borel σ-ring in G that is left G-invariant: µ`(gS) = µ`(S) for measurable set S and g ∈ G. Regularity means that

µ`(S) = inf{ µ`(U) | U ⊃ S, U open } and µ`(S) = sup{ µ`(C) | C ⊂ S, C compact }. Such a measure has the properties that any compact set has ﬁnite measure and any nonempty open set has positive measure. Left invariance of µ` amounts to the property Z Z f(g0g) dµ`(g) = f(g)dµ`(g), ∀ g0 ∈ G, G G for any Haar integrable function f on G. Theorem. ([Halmos], [HR], [L]) If G is a locally compact group, there is a left Haar measure on G, and it is unique up to positive real multiples.

There is also a right Haar measure µr, unique up to positive constant multiples, on G. Right and left Haar measures do not usually coincide. Exercise. Let x y G = | x ∈ ×, y ∈ . 0 1 R R Show that |x|−2dx dy is a left Haar measure on G and |x|−1dx dy is a right Haar measure on G. The (locally compact topological) group G is called unimodular if each left Haar measure is also a right Haar measure. Clearly, G is unimodular if G is abelian. Conjugation −1 by a ﬁxed g0 ∈ G is a homemomorphism of G onto itself, so the measure S 7→ µ`(g0Sg0 ) = −1 µ`(Sg0 )(S measurable) is also a left Haar measure. By uniqueness of left Haar measure, there exists a constant δ(g0) > 0 Z Z −1 f(g0gg0 ) dµ`(g) = δ(g0) f(g) dµ`(g), f integrable G G A quasicharacter of G is a continuous homomorphism from G to C×.

67 Proposition. × (1) The function δ : G → R+ is a quasicharacter (2) δ(g)dµ`(g) is a right Haar measure.

× Proof. The fact that conjugation is an action of G on itself implies that δ : G → R+ is a homomorphism. The proof of continuity is omitted. Note that Z Z Z −1 δ(g0) f(g) dµ`(g) = f(g0 · g0 gg0) dµ`(g) = f(gg0) dµ`(g). G G G

Replacing f by fδ and dividing both sides by δ(g0), we obtain Z Z f(g)δ(g) dµ`(g) = f(gg0)δ(g) dµ`(g). G G

This shows that δ(g)dµ`(g) is right invariant. qed

In view of the above, we may write dµr(g) = δ(g)dµ`(g). The function δ is called the modular quasicharacter of G. Clearly G is unimodular if and only if the modular quasicharacter is trivial. If G is unimodular, we simply refer to Haar measure on G. Exercises: (1) Let dX denote Lebesgue measure on Mn×n(R). This is a Haar measure on Mn×n(R). −n Show that |det(g)| dg is both a left and a right Haar measure on GLn(R). Hence GLn(R) is unimodular.

(2) Let n1 and n2 be positive integers such that n1 + n2 = n. P = P(n1,n2) be the standard parabolic subgroup of GLn(R) corresponding to the partition (n1, n2) (see Chapter 4 for the deﬁnition of standard parabolic subgroup of a general linear group). g1 X Let g = ∈ P , with gj ∈ GLnj (R) and X ∈ Mn1×n2 (R). Let dgj be 0 g2

Haar measure on GLnj (R), and let dX be Haar measure on Mn1×n2 (R). SHow that −n2 −n1 d`g = | det g1| dg1 dg2 dX and drg = | det g2| dg1 dg2 dX are left and right Haar measures on P (respectively). Hence the modular quasicharacter of P is equal to

n1 −n2 δ(g) = | det g1| | det g2| . −1 (3) Show that the homeomorphism g 7→ g turns µ` into a right Haar measure. Conclude R R −1 that if G is unimodular, then G f(g) dµ`(g) = G f(g )dµ`(g) for all measurable functions f.

Proposition. If G is compact, then G is unimodular and µ`(G) < ∞. Proof. Since δ is a continuous homomorphism and G is compact, δ(G) is a compact × × subgroup of R+. But {1} is the only compact subgroup of R+. Haar measure on any locally compact group has the property that any compact subset has ﬁnite measure. Hence

µ`(G) < ∞ whenever G is compact. qed

68 If G is compact, normalized Haar meaure on G is the unique Haar measure µ on G such that µ(G) = 1. When working with compact groups, we will always work relative to R R normalized Haar measure and we will write write G f(g) dg for G f(g) dµ(g).

5.8. Discrete series representations

Let G be a locally compact unimodular topological group. A unitary representation π of G on a Hilbert space V (with inner product h·, ·i) is continuous if for every v, w ∈ V , the function g 7→ hπ(g)v, wi is a continuous function on G. That is, matrix coefficients of π are continuous functions on G. Note that such a representation may be infinite-dimensional. (In particular, if G is a noncompact semisimple Lie group, then all nontrivial irreducible continuous unitary representations of G are infinite-dimensional.) Suppose that (π, V ) is an irreducible continuous unitary representation of G. Let Z be the centre of G. A generalization of Schur’s Lemma to this setting shows that if z ∈ Z, then there exists ω(z) ∈ C× such that π(z) = ω(z)I. Because π is a continuous unitary representation, the function z 7→ ω(z) is a continuous linear character of the group Z. In particular, |ω(z)| = 1 for all z ∈ Z. The representation π is said to be square-integrable mod Z, or to be a discrete series representation, if there exist nonzero vectors v and w ∈ V such that Z |hv, π(g)wi|2 dg× < ∞, G/Z where dg× is Haar measure on the locally compact group G/Z. Thus π is a discrete series representation if some nonzero matrix coefficient of π is square-integrable modulo Z.

Fix an ω as above. Let Cc(G, ω) be the space of continuous functions from f G to C that satisfy f(zg) = ω(z)f(g) for all g ∈ G and z ∈ Z, and are compactly supported modulo Z (there exists a compact subset Cf of G such that the support of f lies inside R × the set Cf Z). Deﬁne an inner product on Cc(G, ω) by (f1, f2) = G/Z f1(g)f2(g) dg . Let 2 1/2 L (G, ω) be the completion of Cc(G, ω) relative to the norm kfk = (f, f) , f ∈ Cc(G, ω). The group G acts by right translation on L2(G, ω), and this deﬁnes a continuous unitary representation of G on the Hilbert space L2(G, ω).

Theorem. (Schur orthogonality relations). Let (π, V ) and (π0,V 0) be irreducible continuous unitary representations of G such that ω = ω0. (1) The following are equivalent: (i) π is square-integrable mod Z. R 2 × (ii) G/Z |hv, π(g)wi| dg < ∞ for all v, w ∈ V . (iii) π is equivalent to a subrepresentation of the right regular representation of G on L2(G, ω).

69 (2) If the conditions of (1) hold, then there exists a number d(π) > 0, called the formal degree of π (depending only on the normalization of Haar measure on G/Z), such that Z × −1 hv1, π(g)w1ihv2, π(g)w2i dg = d(π) hv1, v2ihw1, w2i, ∀ v1, v2, w1, w2 ∈ V. G/Z

(3) If π is not equivalent to π0, then Z hv, π(g)wihv0, π0(g)w0i dg× = 0 ∀ v, w ∈ V, v0, w0 ∈ V 0. G/Z

5.9. Parabolic subgroups and representations of reductive groups The description of the parabolic subgroups of general linear groups and special linear groups over finite fields given in Chapter 4 is valid for general linear and special linear groups over any field F - simply replace the matrix entries in the finite field by matrix entries in the field F . General linear and special linear groups are examples of reductive groups. We do not give the definition of parabolic subgroup for arbitrary reductive groups. Suppose that G is the F -rational points of a connected reductive linear algebraic group, where F = R, F = C, F is a p-adic field (for example, F = Qp), or F is a finite field. The “Philosophy of Cusp Forms” says that the collection of representations of a reductive group G should be partitioned into disjoint subsets in such a way that each subset is attached to an associativity class of parabolic subgroups of G. Two parabolic subgroups P = M n N and P 0 = M 0 n N 0 of G are associate if and only if the Levi factors M and M 0 are conjugate in G. The representations attached to the group G itself are called cuspidal representations, and their matrix coefficients are called cusp forms. If P is a proper parabolic subgroup of G, the representations attached to P are associated to (Weyl group orbits of) cuspidal representations of a Levi factor M of P . (Note that M is itself a reductive group). Furthermore, the representations of G associated to a given cuspidal G 1/2 representation σ of M occur as subquotients of the induced representation IndP (σ ⊗ δP ), where δP is the modular quasicharacter of P (see § 5.7) and σ is extended to a representation of P = M n N by letting it be trivial on N. The problem of understanding the representations of the group G can be approached via the Philosophy of Cusp Forms, and is therefore divided into two parts. The first part is to determine the cuspidal representations of the Levi subgroups M of G, and the second part is to analyze representations parabolically induced from such cuspidal representations. In certain contexts, a cuspidal representation is simply a discrete series representation (see § 5.8 for the definition of discrete series representation). If G is a connected reductive Lie group (for example G = SLn(R) or G = Sp2n(R), then there are two cases to consider.

70 An element of a matrix Lie group is semisimple if it is semsimple as a matrix, that is, it can be diagonalized over the field of complex numbers. A Cartan subgroup of G is a closed subgroup that is a maximal abelian subgroup consisting of semisimple elements. In the first case, G contains no Cartan subgroups that are compact modulo the centre of G (for example, this is the case if G is semisimple and F = C, of if SLn(R) and n ≥ 3), and hence G has no discrete series representations. In the second case, up to conjugacy G contains one Cartan subgroup T that is compact modulo the centre of G, and the discrete series of G are parametrized in a natural way by the so-called regular characters of T . An irreducible unitary representation of G is tempered if it occurs in the decomposition of the regular representation of G on the Hilbert space L2(G) of square-integrable functions on G. If π is a tempered representation of G, then there exists a parabolic subgroup P = M n N and a discrete series representation of M such that π occurs as a constituent G 1/2 of the induced representation IndP (σ ⊗ δP ). If G is a reductive p-adic group (that is, F is a p-adic field), a continuous complex- valued function f on G is a supercusp form if the support of f is compact modulo the centre R of G and N f(gn) dn = 0 for all g ∈ G and all unipotent radicals N of proper parabolic subgroups of G. An irreducible smooth representation (where a smooth representation is as defined in §5.6) of G is supercuspidal if the matrix coefficients of the representation are supercusp forms. Given an irreducible smooth representation π of G, there exists a parabolic subgroup P = M n N of G and a supercuspidal representation σ of M such G 1/2 that π is a subquotient of IndP (σ ⊗ δP ). Hence in this context, it is suitable to interpret “cuspidal representation” as supercuspidal representation. (Recall that a similar result was described in Chapter 4 in the case that F is a finite field). If π is a supercuspidal representation of G, then there exists a quasicharacter ω of the centre Z of G such that π(z) = ω(z)I, z ∈ Z. It is easy to see that if ω is unitary (that is, |ω(z)| = 1 for all z ∈ Z), then π is a discrete series representation. A reductive p-adic group has many supercuspidal representations and hence many discrete series representations. However, there exist discrete series representations that are not supercuspidal.

71 REFERENCES

References for Chapters 1 and 2 Linear Representations of Finite Groups, J.-P. Serre, Graduate Texts in Mathematics, Volume 42, Springer-Verlag. Representations of Finite and Compact Groups, B. Simon, Graduate Studies in Math- ematics, Volume 10, AMS. Representation Theory of Finite Groups, S. Weintraub, Graduate Studies in Mathe- matics, Volume 59, AMS (2003). Representation Theory: A First Course, W. Fulton and J. Harris, Graduate Texts in Mathematics, Volume 129, Springer-Verlag. Introduction to Lie Algebras and Representation Theory, James E. Humphreys, Grad- uate Texts in Mathematics, Volume 9, Springer-Verlag. Character Theory of Finite Groups, Martin Isaacs, Dover. Theory of Group Representations, M.A. Naimark, Springer 1982 (Math Library copy is missing, but Gerstein has a copy). Representation Theory of Finite Groups and Associative Algebras, Curtis and Reiner, Wiley 1962.

References for Chapter 4 J.L. Alperin and R.B. Bell, Groups and Representations, Graduate Texts in Math. 162 (1995), Springer. R. Carter, Finite Groups of Lie Type: Conjugacy Classes and Complex Characters (1985) Wiley. (Note: This book contains an extensive bibliography). P. Deligne and G. Lusztig, Representations of reductive groups over finite fields, Ann. Math. 103 (1976), 103–161. B. Srinivasan, Representations of finite Chevalley groups, Lecture Notes in Math. 764 (1979) Springer.

References for Chapter 5 and 6 [B] Daniel Bump, Lie Groups, Graduate Texts in Mathematics 225, 2004, Springer. [C] W. Casselman, Introduction to the Theory of Admissible Representations of Reductive p-adic Groups, preprint, 1974. Available at www.math.ubc.ca/ cass/reaserch.html.

72 [Hall] Brian C. Hall, Lie Groups, Lie Algebras, and Representations, An Elementary Intro- duction, Graduate Texts in Mathematics 222, 2003, Springer. [Halmos] P. Halmos, Measure Theory, D. Van Nostrand Company, Inc. New York, 1950. [HR] E. Hewitt and K. Ross, Abstract Harmonic Analysis. Vol. I, Structure of topological groups, integration, theory, group representations volume 115 of Grundlehren der Mathematischen Wissenshaften, second edition, 1979, Springer-Verlag. [Hu] T. Husain, Introduction to Topological Groups, W.B. Saunders Co., 1966. [K1] Anthony W. Knapp, Representation Theory of Semisimple Groups: An Overview Based on Examples, Princeton University Press, 1986. [K2] Anthony W. Knapp, Lie Groups: Beyond an Introduction, Birkhauser 1996. [L] L. Loomis, An Introduction to Abstract Harmonic Analysis, D. Van Nostrand Com- pany, Inc., 1953. [M] F. Murnaghan, Mat 1197 course notes: Representations of reductive p-adic groups, available at www.math.toronto.edu/ murnaghan. [P] L.S. Pontryagin, Topological Groups, second edition, Gordon and Breach, 1966.